HEAT  ENGINEERING 


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HEAT  ENGINEERING 


A  TEXT  BOOK  OF  APPLIED  THERMODYNAMICS 

FOR  ENGINEERS  AND  STUDENTS 

IN  TECHNICAL  SCHOOLS 


BY 
ARTHUR  M.  GREENE,  JR. 

PROFESSOR    OF    MECHANICAL    ENGINEERING,    RUSSELL     SAGE    FOUNDATION 

RENSSELAER     POLYTECHNIC     INSTITUTE;     SOMETIME     JUNIOR 

DEAN,    SCHOOL  OF   ENGINEERING,    UNIVERSITY 

OF   MISSOURI 


FIRST  EDITION 


McGRAW-HILL  BOOK  COMPANY,  INC. 

239  WEST  39TH  STREET,  NEW  YORK 

6  BOUVERIE  STREET,  LONDON,  E.  C. 

1915 


COPYRIGHT,  1915,  BY  THE 
MCGRAW-HILL  BOOK  COMPANY,  INC. 


THE    MAPI.E     I'KKSS     Y  O  H  K    PA 


PREFACE 

For  many  years  the  author  has  giyen  lectures  supplementing 
the  text-books  used  as  a  basis  for  a  course  in  heat  engineering. 
His  aim  in  preparing  this  book  has  been  to  bring  together  his 
various  notes  with  statements  of  the  investigations  and  writings 
of  others  to  make  a  complete  treatment  of  the  important  phases 
of  this  subject.  In  doing  this  he  has  given  credit  to  the  authors 
and  investigators  quoted.  Certain  of  the  original  sources  have 
been  quoted  so  that  the  student  may  learn  the  use  of  references. 
It  is  hoped  that  many  studying  this  book  will  refer  to  these 
original  papers. 

The  work  presupposes  a  course  in  theoretical  thermodynamics 
such  as  that  given  in  the  treatises  of  Wood,  Peabody  or  Goode- 
nough.  Because  of  the  difference  in  symbols,  nomenclature  or 
point  of  view  of  various  authors  and  to  serve  for  reference  or 
for  the  derivation  of  formulae  used  in  the  text,  the  first  chapter  of 
this  book  has  been  written.  It  is  not  intended  that  this  chapter 
shall  be  used  as  a  part  of  the  course  for  it  is  an  outline  only  of  the 
thermodynamic  theory.  It  should  be  used  to  give  a  review  of  the 
subject  or  as  a  basis  for  the  formulae  used.  In  shaping  this  chapter 
the  author  has  been  guided  by  his  experience  in  teaching  this 
subject  from  many  texts.  The  treatment  of  availability  and 
entropy  has  been  based  on  the  excellent  work  on  thermodynamics 
by  Goodenough. 

Numerical  problems  have  been  solved  at  various  points  in  the 
text  to  illustrate  the  principles  of  the  subject  and  to  apply  them  to 
actual  engineering  work:  The  problems  have  been  solved  in 
detail  to  give  the  student  one  manner  of  attack  as  well  as  an  order 
for  the  arrangement  of  computations  for  clearness.  Unless  the 
student  can  apply  the  various  formulae  and  theories  he  has  failed 
to  attain  that  for  which  this  book  was  written.  In  addition  to  the 
problems  and  solutions  a  series  of  questions  on  the  various  topics 
of  the  text  and  a  set  of  problems  illustrating  their  use  have  been 
placed  at  the  end  of  each  chapter.  These  may  be  used  by  the 
student  in  preparation  of  an  assignment  or  by  the  teacher  for 
blackboard  recitations. 


342912 


vi  PREFACE 

The  author  not  only  expresses  his  thanks  to  those  whose  works 
he  has  used  and  whose  names  he  has  placed  in  the  first  part  of  the 
index  but  to  those  whose  writings  he  has  studied  as  a  student  and 
teacher  and  whose  work  or  whose  view  point  he  has  absorbed. 
He  especially  thanks  his  wife,  Mary  E.  Lewis  Greene,  for  her  aid  in 
the  preparation  of  manuscript,  proof  and  final  arrangement  of 
work.  A.  M.  G.,  Jr. 

SUNNYSLOPE,  TROY,  N.  Y., 
February  22,  1915. 


CONTENTS 

PAGE 

PREFACE    v 

LIST  OF  SYMBOLS      xi 

CHAPTER  I 

FUNDAMENTAL  THERMODYNAMICS 1 

Availability — Graphical  Representation  of  Heat  on  P.  V.  Plane — 
Scale  of  Temperature — Carnot  Cycle — Reversed  Cycle — Maxi- 
mum Efficiency — Conditions  for  Availability  of  Heat — Second 
Law — Loss  of  Availability — Entropy — Entropy  around  a  Cycle — 
Entropy  Diagram — Characteristic  Equations — Heat  on  a  Path — 
Differential  Equations  and  Relations — Perfect  Gases — Differen- 
tial Heat  Equations  for  Perfect  Gases — Specific  Heats — Isothermal 
and  Isodynamic  Lines — Adiabatic — Thermodynamic  Lines — 
Equality  of  Gas  Scale  and  Kelvin  Scale — Heat  Content  of  Gases — 
Entropy  of  Gases — Cycles  and  Cross  Products — Saturated  Steam 
and  Other  Vapors — Intrinsic  Energy — Heat  on  a  Path — Heat 
Content — Entropy — Differential  Effect  of  Heat — Superheated 
Vapor — Specific  Heats  at  Constant  Pressure  and  k's — T-S  and 
I-S  Charts — Heat  on  Paths — Flow  of  Fluids — Throttling  Action — 
Velocity  of  Various  Substances — Discharge  from  Orifices. 

CHAPTER  II 

HEAT  ENGINES  AND  EFFICIENCIES 62 

Method  of  Reporting  Performance  of  Engines — High-speed  Non- 
condensing  Engine — High-speed  Non-condensing  Compound 
Engine — Locomobile  Engine — Pumping  Engine — Steam  Turbine — 
Producer  Gas  Engine — Blast-furnace  Gas  Engine — Diesel  Oil 
Engine — Results  of  Various  Tests — Topics — Problems. 

CHAPTER  III 

HEAT  TRANSMISSION 72 

Laws  of  Transmission — Mean  Temperature  Difference — Deter- 
mination of  K-Steam  Condensers — Ammonia  Condensers — Brine 
Coils — Evaporators  and  Feed-water  Heaters — Heat  from  Liquids 
to  Liquids — Factor  of  Safety — Radiators — Heat  through  Walls 
and  Partitions — Efficiency  of  Heat  Transmissions — Topics — 
Problems. 

vii 


viii  CONTENTS 


CHAPTER  IV 

PAGE 

Am  COMPRESSORS 118 

Various  Types — Work  of  Compression — Effect  of  Clearance — 
Effect  of  Leakage — Volumetric  Efficiency — Horse-power  and 
Power  of  Motor — Temperature  at  the  end  of  Compression — Com- 
pression Curve — Heat  Removed  by  Jacket — Saving  due  to  Jacket — 
Water  Required  for  Jacket — Multistaging  and  Intermediate  Pres- 
sures— Intermediate  Temperatures — Heat  Removed  by  Inter- 
cooler — Amount  of  Water  for  Intercooler — Water  Removed  in 
Intercooler — Effect  of  Leakage  in  Multistage  Compressions — Dis- 
placement of  Cylinders — Size  of  Inlet  and  Outlet  Pipes  and 
Valves — Percentage  Saving  due  to  Multistaging  over  a  Single 
Stage — Unavoidable  Loss  of  Compression  on  Two  Stages — 
Work  on  Air  Engine — Loss  due  to  Cooling  after  Compression — 
Unavoidable  Loss  due  to  Expansion  Line — Loss  of  Pressure  in 
Pipe  Line — Loss  due  to  Leakage  from  Transmission  Line — Loss 
due  to  Throttling — Gain  from  Preheating — Power  and  Displace- 
ment of  Air  Engine  or  Motor — Fan  Blowers — Governing — Motors 
— Losses  in  Transmission — Logarithmic  Diagrams — Table  of 
Power — Topics — Problems . 


CHAPTER  V 

THE  STEAM  ENGINE 167 

Action  of  Steam  Engine — Cycle  of  Steam  Engine — Heat  and  Effi- 
ciency of  Cycle — Steam  Consumption — Temperature-Entropy 
Diagrams — Effects  of  Changes — Calorimeters — Analysis  of  Tests — 
Hirn's  Analysis — Temperature  Entropy  Analysis — Missing  Quan- 
tity and  Initial  Condensation — Experiments  of  Effect  of  Cylinder 
Walls — Valve  Leakage — Steam  Consumption  by  Clayton's  Method 
— Values  of  N  for  Expansion  Lines — Expansion  Lines — Use  of  Rec- 
tangular Hyperbola — Construction  of  Expansion  Curves — Mean 
Effective  Pressure — Real  and  Apparent  Ratio  of  Expansion — 
Stumpf  Engine — Size  of  Engines — Best  Point  of  Compression — 
Speed — Locomobile— Topics — Problems. 


CHAPTER  VI 

MULTIPLE  EXPANSION  ENGINES 240 

Action — Combined  Cards — Computation  of  Cards  for  Construction 
— Equivalent  Work  done  by  One  Cylinder — Determination  of 
Relative  Sizes  of  Cylinders — Jacketing — Reheaters — Governing 
— Bleeding  Engines  or  Turbines — Regenerative  Engines — Testing 
and  Analysis — Binary  Engines — Topics — Problems. 


CONTENTS  ix 

CHAPTER  VII 

PAGE 

STEAM  NOZZLES,  INJECTORS,  STEAM  TURBINES 265 

Nozzles — Injectors — Steam  Nozzle  Velocity — Water  Velocities — 
Theory  of  the  Injector — Heat  Equation  of  the  Injector — Steam 
Weight — Steam  Nozzle — Combining  Tube — Delivery  Tube — 
Density — Impact  Coefficient — Injector  Details — Steam  Turbines 
— Action  of  Jets — Maximum  Efficiency — Types  of  Turbines — 
Efficiency — Similarity  of  Action  of  Turbine  and  Engine — Compu- 
tations and  Design — Reheat  Factor — Combined  Engine  and  Tur- 
bine— Allowance  for  Change  of  Running  Conditions — Combined 
Engine  and  Turbine — Topics — Problems. 

CHAPTER  VIII 

CONDENSERS,  COOLING  TOWERS  AND  EVAPORATORS 333 

Types  of  Condensers — Pressures  in  a  Condenser — Condenser 
Design — Problems — Cooling  Towers — Size  of  Tower  and  Mats — 
Spray  Nozzles  and  Ponds — Accumulators  and  Evaporators — 
Double  Bottoms — Temperature  of  Boiling — Topics — Problems. 

CHAPTER  IX 

INTERNAL  COMBUSTION  ENGINES  AND  COMBUSTION 359 

Development  of  Engine — Otto  Cycle — Air  Standard  Efficiency — 
Atkinson  Cycle  and  Diesel  Cycle — Actual  Engines — Governing — 
Ignition — Heat  Transfer  to  Walls — Fuels — Combustion  of  Fuels — 
Gas  Composition — Problem — Temperatures  at  Corners  of  Card — 
Adiabatics — Efficiency  Change — Temperature  after  Explosion — 
Temperature  Entropy  Diagram — Logarithmic  Diagram — Test  of 
Gas  Engine — Combustion  of  Fuels  for  Boilers — Surface  Combus- 
tion— Topics — Problems. 

CHAPTER  X 

REFRIGERATION 411 

Air  Machines — Problem — Effect  of  Clearance  and  Friction  Mois- 
ture— Vapor  Refrigerating  Machines — Effect  of  Substances — Ab- 
sorption Apparatus — Properties  of  Aqua  Ammonia — Investigation 
of  Absorption  Apparatus — Topics — Problems. 

INDEX  .   455 


SYMBOLS  USED  IN  TEXT 

A  =  -j  =  77g  =  const,  to  change  foot  pounds  to  B.t.u. 

A  =  constant  in  Reynold's  equation. 

a  =  cleaness  factor  Orrok's  formula. 

a  =  coefficient  of  heat  transmission  at  surface. 

a  =  coefficient  in  various  equations  or  length  of  line. 

pV 

B  =  gas  constant  =  j^  =  foot-pounds  per  deg.  per  pound. 

B  =  constant  in  Reynold's  equation. 

b  =  constant  in  various  equations  or  lengths  of  lines. 

6  =  fraction  of  volume  at  compression  in  Mark's  formula. 

b.h.p.  =  brake  horse-power. 

B.t.u.  =  British  thermal  units. 

C  =  constant  in  equation. 

c  =  coefficient  of  conduction  or  heat  transmission. 

c  =  specific  heat  of  water  or  length  of  line. 

c'  =  specific  heat  of  liquid. 

c"  =  specific  heat  of  saturated  steam 

CP  =  specific  heat  at  constant  pressure  of  1  cu.  ft. 

cp  =  specific  heat  at  constant  pressure. 

Cv  =  specific  heat  of  constant  volume  of  1  cu.  ft. 

cv  =  specific  heat  of  constant  volume. 

D  =  displacement  of  cylinder  in  cubic  feet  or  length  of  card  in  inches. 

d  =  diameter  of  cylinder  or  pipe  in  feet  (  or  inches). 

d'  =  hydraulic  radius. 

d  =  constant  in  equation  for  air  movement  in  heat  transmission  or 

length  of  line. 

D.H.P.  =  delivered  horse-power. 

e  =  area  of  elementary  strip  in  Kelvin  diagram. 

e  =  fraction  of  steam  at  cut-off. 

e  =  weight  of  evaporation  in  evaporator. 

e  =  constant  for  material  in  formula  for  heat  tra-nsmission. 

F  =  area  or  surface  in  square  feet  (or  square  inches) 

F  =  Maxwell's  thermodynamic  potential. 

/  =  leakage  factor,  diagram  factor,  or  percentage  friction. 

/  =  \/l  —  y  =  friction  factor  for  velocity. 

G  =  Ibs.  of  water  per  minute  or  per  pound  of  steam. 

g  =  acceleration  of  gravity. 

H  =  heat  from  internal  friction  in  B.t.u. 

H  =  heat  in  1  cu.  ft.  of  gas. 

h  =  feet  head. 

h'  =  lift  of  injector  in  feet. 

H.P.  =  horse-power. 

/  =  heat  content  of  M  Ibs.  of  substance. 

xi 


xii  SYMBOLS   USED  IN  TEXT 

i  =  heat  content  of  1  Ib.  of  substance. 

i.h.p.  =  indicated  horse-power. 

J  =  Joule's  equivalent  =  778  =  constant  to  change  B.t.u.  to  foot- 
pounds. 

K  =  coefficient  of  conduction  and  coefficient  of  various  equations. 

k  =  ratio  of  cp  to  cv  or  coefficient  of  equation. 

KZ  =  clearance  factor. 

Kw  =  kilowatts. 

L  =  stroke  in  feet. 

I  =  percentage  clearance  or  periphery  of  valves. 

I  =  length  of  path  or  thickness  of  material  in  feet. 

IP  and  lv  =  latent  heats. 

M  =  mass  of  substance  in  pounds. 

m  —  mass  of  1  cu.  ft.  of  substance  in  pounds  or  constant  in  equations. 

Ma  =  apparent  weight  of  steam  per  card,  indicated. 

Mi  =  apparent  weight  of  steam  per  card,  indicated. 

m/  =  weight  of  steam  per  cubic  foot  of  displacement. 

Mm  —  missing  weight  of  steam. 

M'o  =  clearance  steam  per  card. 

m0  =  weight,  of  cubic  feet  of  steam. 

m.q.  =  missing  quantity  in  per  cent,  of  steam  used. 

M  total  =  steam  used  by  engine. 

N  =  revolutions  per  minute. 

n  =  exponent  of  volume  terms  in  equations. 

n  =  no.  of  molecules. 

n  =  heat  of  expansion. 

o  =  heat  of  pressure  change. 

P  =  total  force  or  pressure. 

P  =  mean  effective  pressure  on  piston. 

p  =  normal  pressure  in  pounds  per  square  foot  (pounds  per  square 
inch  or  millimeter  of  mercury). 

pm  —  friction  factor  in  pipes. 

Q  =  heat  added  to  M  Ibs.  in  B.t.u. 

q  =  heat  per  pound  of  substance  or  heat  of  solution  of  aqua  ammonia. 

q'  =  heat  of  liquid,  at  outlet,  q'0;  at  inlet  q'i. 

q"  =  total  heat  in  1  Ib.  of  saturated  steam. 

q'"  =  total  heat  of  1  Ib.  of  superheated  steam. 

R  =  universal  gas  constant  =  1544  ft.-lbs.  per  deg.  per  molecule. 

R  =  total  ratio  of  expansion. 

r  =  heat  of  vaporization. 

r  =  ratio  of  expansion. 

=  cut-off. 

TTI  =  entropy  of  vaporization. 

Tr  =  real  ratio  of  expansion. 

r.e.  =  residual  energy. 

R.H.  =  reheat  factor. 

s  =  stroke  in  feet. 

s  =  normal  cylinder  surface  per  cubic  foot  of  displacement. 


SYMBOLS   USED  IN  TEXT  xiii 

s'  =  Entropy  of  1  Ib.  liquid. 

s"  =  total  entropy  of  1  Ib.  steam. 

S2  —Si  =  change  of  entropy  for  M  Ibs.  of  substance. 

82—81    =  change  of  entropy  per  pound  of  substance. 

T  =  temperature  in  degrees  from  absolute  zero. 

T  =  temperature  difference  in  Heck's  formula. 

t  =  temperature  from  F.  or  C.  zero. 

Ttat  =  temperature  of  boiling. 

Tsoi  =  temperature  of  boiling  of  aqua  ammonia. 

U       .      =  intrinsic  energy  of  M  Ibs.  of  substance  in  foot-pounds. 

u  =  intrinsic  energy  of  1  Ib.  of  substance. 

V  =  volume  of  M  Ib's.  of  substance  in  cubic  feet. 

v  =  volume  of  1  Ib.  of  substance  in  cubic  feet. 

v'  =  volume  of  1  Ib.  of  liquid. 

v"  =  volume  of  1  Ib.  of  dry  saturated  steam. 

W  =  work  in  foot-pounds  for  M  Ibs.  of  substance. 

w  =  work  for  1  Ib.  of  substance. 

w  =  velocity  of  substance  in  feet  per  second. 

wa  =  actual  velocity  of  jet  in  feet  per  second. 

Wb  =  velocity  of  blade  in  feet  per  second. 

WT  =  relative  velocity  of  jet. 

wt.  =  molecular  weight. 

x  =  quality  or  dryness  factor  =  pounds  of  steam   per   pound    of 

mixture. 

x  =  per  cent,  of  NH3  in  1  Ib.  of  solution. 

x  =  point  of  compression,  distance  moved  and  variable  length  of 

injector. 

x"  =  distance  on  card  occupied  by  boiler  steam. 

y  =  friction  factor  for  velocity. 

y  =  amount  of  liquor  to  produce  given  change  in  concentration. 

z  =  pounds  of  water  per  pound  of  steam  in  injector. 

a  =  acceleration. 

a.  =  temperature  coefficient. 

a  =  angle  of  actual  velocity. 

0  =  angle  of  relative  velocity. 

7  =  angle  on  I-S  diagram. 

5  =  angle  on  7-5  diagram. 

A  =  relative  density  in  injector. 

A£  =  temperature  difference. 

«  =  base  of  natural  logarithms. 

-n  =  efficiency  or  over-all  efficiency. 

in  =  Carnot  efficiency. 

772  =  type  efficiency. 

rj3  =  theoretical  efficiency. 

774  =  practical  efficiency. 

775  =  actual  efficiency. 

"n o  =  mechanical  efficiency. 

ije  =  electrical  efficiency. 

77*  =  kinetic  efficiency. 


xiv  SYMBOLS  USED  IN  TEXT 

rjm  =  mechanical  efficiency. 

rjn  =  nozzle  efficiency. 

ijr  =  refrigerative  efficiency. 

77S  =  efficiency  of  stage. 

ijt  =  over-all  efficiency. 

iffw  =  efficiency  of  weight. 

6  =  temperature  of  material. 

X  =  coefficient  of  conduction  of  gas. 

At  =  materials  factor  in  Orrok's  formula. 

p  =  internal  heat  of  vaporization. 

P  =  relative  humidity. 

P  =  steam  richness  factor  in  Orrok's  formula. 

<f>  =  Maxwell's  thermodynamic  potential. 

<f>  =  coefficient  in  Nicolson's  formula  of  heat  transmission. 

^  =  external  work  in  making  steam. 


HEAT  ENGINEERING 

CHAPTER  I 
FUNDAMENTAL  THERMODYNAMICS 

Heat  is  a  form  of  energy  and  as  such  it  may  be  measured  in 
any  unit  of  energy.  The  customary  unit  is  the  British  thermal 
unit,  B.t.u.,  although  heat  may  be  measured  in  foot-pounds,  ergs, 
joules,  calories,  horse-power  hours,  kilowatt  hours  or  any  other 
unit  of  energy.  The  B.t.u.  is  Jf  80  of  the  amount  of  heat  neces- 
sary to  raise  the  temperature  of  1  Ib.  of  water  from  32°  F.  to 
212°  F. 

Heat  energy  is  a  form  of  energy  due  to  a  vibration  or  motion 
of  the  molecules  of  a  body.  .It  may  be  produced  by  the  transfor- 
mation of  other  forms  of  energy  into  heat,  as  when  mechanical 
energy  of  a  moving  train  is  changed  into  heat  by  brakes. 
When  other  forms  of  energy  are  produced  from  heat  energy,  how- 
ever, it  is  found  that  only  a  portion  of  the  heat  energy  may  be 
transformed.  This  leads  at  once  to  the  separation  of  energy 
into  two  classes:  high-grade  energy  and  low-grade  energy. 
High-grade  energy  is  any  kind  of  energy  which  may  be  completely 
changed  into  any  other  form.  Mechanical  and  electrical  energy 
are  examples  of  this  while  heat  energy  is  low  grade  since  it  cannot 
be  changed  completely  into  any  other  form  of  energy.  For  this 
reason  all  of  the  heat  energy  in  a  body  is  not  available  for  trans^ 
formation.  The  fractional  part  of  the  heat  energy  which  is 
available  for  transformation  into  another  form  is  known  as  the 
availability  of  heat  energy.  If  Q  heat  units  are  in  a  system 
and  if  part  of  these  are  changed  into  another  form  of  energy  and 
Qf  heat  units  remain  in  the  system,  then  Q  —  Q'  heat  units  must 
have  been  changed.  If  this  is  all  that  could  have  been  changed. 

Q-Q' 

— Q —  represents  the  fractional  part  or  the  availability.     It  has 

been  shown  by  Joule  and  others  that  when  mechanical  work  is 
changed  into  heat  or  when  heat  is  partially  changed  into  mechan- 
ical energy  there  is  a  definite  relation  between  the  work  and  the 

1 


2  HEAT  ENGINEERING 

heat  in  the  first  case  and  between  the  disappearance  of  heat  and 
the  work  produced  thereby  in  the  second  case.  This  statement 
is  known  as  the  first  law  of  thermodynamics.  The  numerical 
relation  between  heat  and  work  is  known  as  Joule's  equivalent, 
J.  It  is  equal  to  778  ft.-lbs. 

1  B.t.u.  =  778  ft.-lbs.  (1) 

JQ  =  W  (2) 

For  convenience  the  reciprocal  of  'J  is  used  at  times,  the  symbol 
for  this  being  A. 

7  =  A  W 

Q  =  AW  (4) 

Q  =  Heat  in  B.t.u.         W  =  Work  in  ft.-lbs. 

In  French  units  the  value  of  J  is  426  kg.  m.  =  1  kg.  deg.  calorie. 
Now  when  heat  is  added  to  a  body  it  may  increase  the  internal 
energy  of  that  body  and  do  external  work  or 

JdQ  =  dU  +  dW  (5) 

This  formula  is  a  mathematical  statement  of  the  first  law  of 
thermodynamics  which  is  only  one  aspect  of  the  law  of  the  con- 
servation of  energy ;  heat  added  to  a  body  increases  the  energy 
of  that  body  and  does  external  work.  U  is  the  symbol  used  to 
indicate  internal  energy  or  intrinsic  energy  of  a  body.  The  capi- 
tal letters  Q,  U,  W,  refer  to  a  body  of  any  weight.  If  a  body  of 
unit  weight  is  considered,  small  letters  are  used. 

Jdq  =  du  +  dw  (6) 

Now  in  the  above  equation  any  one  of  the  differential  quanti- 
ties may  be  zero  or  have  any  sign.  Thus  if  dq  =  0,  no  heat  is 
added  and  such  action  is  known  as  adiabatic  action.  If  du  =  0, 
there  is  no  change  in  intrinsic  energy  and  the  action  is  known  as 
isodynamic  action,  while  if  dw  =  0,  no  external  work  is  done,  and, 
as  will  be  seen  later,  there  is  no  change  in  volume.  The  signs 
of  these  terms  may  be  anything  and  the  equation  is  true.  This 
equation  is  one  of  the  important  ones  in  thermodynamics.  If  at 
any  time  the  change  of  intrinsic  energy  is  known  on  any  path 
or  during  any  change  of  state  and  if  the  work  is  known  during 
that  change,  then  the  sum  of  these  two  will  be  the  heat  required 
during  this  change.  If  positive,  heat  must  be  added ;  if  negative, 
heat  must  be  abstracted. 


FUNDAMENTAL  THERMODYNAMICS  3 

Now  p  represents  the  pressure  on  unit  area  (1  sq.  ft.)  of  the 
surface  of  the  body  considered  and  it  is  assumed  uniform  and  nor- 
mal to  the  surface  surrounding  the  body.  If  then  the  total  area 
of  the  surface  of  the  body  isF,pF  is  the  total  normal  force  on  the 
body.  If  this  surface  is  moved  through  the  normal  distance  dx, 
the  work  done  is 

pFdx  =  dW 
but  since 

Fdx  =  dV 

pdV  =  dW  (7) 

or 

pdv  =  dw 

The  equation  (6)  may  therefore  be  written 

Jdq  =  du  +  pdv  (8) 

In  this  equation  du  must  be  measured  in  the  same  units  as  pdv, 
namely  foot-pounds,  and  dq  on  account  of  the  constant  J  must  be 
measured  in  heat  units.  The  total  heat  is  the  integral  of  dq  or 


The  internal  energy  of  a  body  depends  on  the  condition  of  the 
body  and  not  on  the  method  of  bringing  it  to  that  state  and  conse- 
quently the  change  in  intrinsic  energy,  which  is  I  du,  does  not 

depend  on  the  manner  in  which  one  state  is  changed  to  the  other 
but  merely  on  the  states  at  the  beginning  and  the  end  of  the 
operation.  Hence,  if  the  intrinsic  energy  at  state  a  is  ua  and  at 
state  b  is  Ub,  the  value  of  the  integral  du  between  these  two 
states  is  Ub  —  ua.  This  is  true  whatever  the  path  may  be. 

•b 

pdv  represents  the  area  beneath  a  curve  on  the  pv  plane  and 


1 


this  area  depends  on  the  path  considered.  Hence  the  value  of 
this  integral  can  only  be  told  after  the  curve  or  path  is  known. 
The  integral  of  dq  depends  therefore  on  the  path  since,  although 

I  du  is  independent  of  the  path,    I  dw  or    I  pdv  does  depend 

on  the  path.  A  differential  whose  integral  does  not  depend  on 
the  path  is  called  an  exact  differential  because  it  can  be  inte- 


4  HEAT  ENGINEERING 

grated  directly.  Hence  there  must  be  some  functional  relation 
between  the  independent  variables  in  regard  to  which  it  is  being 
integrated,  otherwise  a  path  would  have  to  be  fixed  or  some  rela- 
tion would  have  to  be  known  to  give  one  variable  in  terms  of  the 
other  to  make  the  integration  possible.  Thus 


Cydx  =      CdX 


is  not  an  exact  differential  since  there  must  be  some  known  rela- 
tion between  y  and  x  before  the  integration  can  be  performed. 

Now    I  dx  =  x  and    I  (xdy  +  ydx)  =  xy 

are  each  directly  integrable  and  their  definite  integral  values 
depend  only  on  the  limits  and  not  on  the  path  between  the  limits 
and  for  this  reason  each  quantity  behind  the  integral  sign  is  an 
exact  differential.  It  is  seen  that  if  exact  differentials  be  inte- 
grated around  a  closed  path  their  value  would  be  zero  since  both 
limits  are  the  same.  This  is  one  way  of  telling  whether  or  not  a 
differential  is  exact.  If  the  integral  on  every  closed  path  is 
zero,  the  differential  is  exact.  There  is  one  other  criterion  by 
which  an  exact  differential  may  be  told  or  a  relation  which  must 
exist  if  a  differential  is  exact,  and  if  known  to  be  a  function  of 
two  independent  variables.  Thus  if 

dX  =  Mdx  +  Ndy 
dX  is  an  exact  differential  if 

181      ?-?  <»> 

by         dx 

If  however  dX  is  known  to  be  exact  by  some  other  method, 
then 

dM  .  dN 

-r—  must  equal  —  (9 ) 

oy  ox 

The  reason  for  this  is  seen  from  the  fact  that  if  dX  is  exact  there 
must  be  some  functional  relation 

X  -  J(xy) 
hence 

8X  j      ,    8X  j 
dx  =       "  dx  +       - 


now 


but 


hence 


FUNDAMENTAL  THERMODYNAMICS 


dX  dX 

-T—  =  M  and  -r—  =  N 
8x  dy 


dxdy        dydx 


_  dM  =  d2X    _  dN_ 

dxdy  ~   dy     ~  dydx  ~  dx 


This  relation  between  the  differential  coefficients  M  and  N 
may  be  used  to  determine  whether  or  not  the  differential  is  exact, 
or  if  known  to  be  exact,  the  equality  of  the  partial  derivative  of 
M  and  N  may  be  used  to  determine  new  relations  as  will  be  seen 
later. 

The  fundamental  equation  may  be  written 


Jq 


—  ua+    I  pdv 


(10) 


If  q  =  0,  the  line  is  an  adiabatic  and 


r 

—  Ub   =      \p 

Ja 


=    I  pdv 


(11) 


This  means  that  work  during  an  adiabatic  change  is  equal 
to  the  change  of  intrinsic  energy,  or  work  is  done  at  the  expense 
of  intrinsic  energy.  If  ab  °°  , 
Fig.  1,  represents  an  adiabatic 
path  of  a  substance  'on  the  pv 
plane,  the  area  la&2  represents 
the  change  of  intrinsic  energy 
from  a  to  b.  If  this  is  car- 
ried out  to  infinity  there  can 
be  no  further  area  under  the 
curve  and  the  area  from  a  to 

this  point  must  represent  the  i  v       2 

total  intrinsic  energy  at  a  since      FIG.  1.— Adiabatic  on  the  pV  plane, 
no  heat  has  been  added  and 

work  has  been  done  to  the  point  of  zero  temperature.  Although 
this  area  is  infinite  in  extent  it  is  not  infinite  in  value  since  the 
amount  of  energy  in  a  body  cannot  be  infinite.  This  is  seen  to 
be  true  mathematically  since  the  curve  approaches  the  v  axis  rap- 


6 


HEAT  ENGINEERING 


idly  and  the  height  of  the  curve  is  practically  zero  after  a  short 
distance  to  the  right. 

If  u  does  not  change  on  the  curve  (the  isodynamic)  the  heat 
added  is  equal  to  the  work  done. 


Jq 


r 

=    I  pdv 

J  a 


(12) 


If  there  is  no  work,    I  pdv  =  0,  or  dv  =  0,  and  v  =  constant. 

In  this  case 

Jq  =  ub  —  ua  (13) 

GRAPHICAL  REPRESENTATION  OF  HEAT  ON  P.V.  PLANE 

In  Fig.  2  let  ab  be  any  path.  Draw  from  a  and  b  two  adia- 
batics  to  infinity  and  from  a  draw  the  isodynamic  until  it  strikes 
the  adiabatic  from  b  at  c. 


12  v  3 

FIG.  2. — Graphical  representation  of  heat  added. 


Now    la  oo   =  u 
26  oo    =  Ub 


-I- 

Jo, 


Ia62  =    I  pdv 

J  a 

r 

26  oo    -f  la62  =   Ia6  oo    =  Ub  +  |  pdv 
Ia6  oo  —  la  °o   =  oo  a6  ro   =  u 


r 

I  pdz;  —  ua  — 

*)  a 

b  —  ua+  I  pdz;  = 

J  a 


Jq 


FUNDAMENTAL  THERMODYNAMICS  7 

This  gives  the  important  relation  that  the  area  on  the  pv 
plane  between  any  path  and  the  two  adiabatics  from  the  ex- 
tremities of  the  path  to  infinity  is  equal  to  the  heat  added  on  the 
path. 

This  area  is  infinite  in  extent  and  consequently  cannot  be 
represented  graphically.  To  make  it  finite  and  definite,  the  iso- 
dynamic  ac  was  drawn.  Now  ua  =  uc.  Hence  3c  °°  =  la  c°  and 

lab  co  —  3coo   =  Jq  =  Ia6c3 

or  the  area  beneath  any  path  added  to  that  beneath  an  adiabatic 
from  the  second  point  of  that  path  to  the  intersection  of  the 
adiabatic  and  the  isodynamic  from  the  first  point  is  equal  to  the 
heat  added  on  the  path.  It  will  be  seen  that 


Iab2 
and 


r 

=    I  pdv 

J  a 


26c3    =  Ub   —  Uc   =  Ub   —  Ua 

This  latter  statement  should  be  clear  since  uc  =  ua. 

SCALE  OF  TEMPERATURE 

When  heat  is  added  to  a  body  this  body  comes  into  a  state 
in  which  it  will  transmit  heat  to  another  body  with  which  it  had 
been  previously  in  contact.  This  ability  to  transmit  heat  to 
another  body  is  determined  by  a  property  which  is  termed  tem- 
perature. The  temperature  is  measured  by  allowing  an  instru- 
ment to  come  into  thermal  equilibrium  with  the  body  and  then 
noting  the  effect  on  the  instrument.  If  one  body  is  at  a  higher 
temperature  than  another  it  will  transmit  heat  to  that  body. 

Suppose  ab  becomes  a  line  of  constant  temperature  T,  such  a 
line  is  called  an  isothermal  (Fig.  3).  Ifa<»  and  b*>  are  adia- 
batics, the  area  <x>ab<*  is  the  heat  added  on  this  line  ab  and  is 
finite  in  value  since  it  is  equal  to  Iab2  +  26  oo  —  la°o  and  each 
of  these  is  finite.  If  qab  is  divided  by  T  the  result 

~  may  be  called  e. 

If  the  isothermal  a'b'  is  drawn  below  ab  so  that  the  area  aWar 
is  equal  to  e  and  then  a  second  isothermal  a"b"  is  so  placed  that 
is  equal  to  e  and  this  is  continued,  it  is  found  that  there 


8 


HEAT  ENGINEERING 


will  be  T  such  isothermals  drawn.  Each  isothermal  is  so  drawn 
that  the  small  areas  are  equal  and  these  isothermals  determine 
definite  temperatures.  These  temperatures  form  a  scale  and 
since  this  results  from  considerations  of  absolute  units  of  work 
and  does  not  depend  on  any  particular  substance  it  is  called 
the  Kelvin  absolute  scale  of  temperature. 

The  ordinary  scale  of  temperature  was  first  determined  by  the 
effects  of  heat  on  mercury  or  other  fluids  but  on  account  of  changes 
in  the  rate  of  expansion  of  these  fluids  by  which  the  temperatures 
were  measured,  gases  were  suggested  as  the  media  for  measure- 
ment and  hydrogen  was  found  to  be  the  most  constant  for  all 
measurements.  There  are  two  scales  in  common  use  in  this 
country,  the  Fahrenheit  in  which  the  temperature  of  melting  ice 
is  32°  and  that  of  boiling  water  under  standard  conditions  is 


I       V     2 
FIG.  3. — Isothermals  for  Kelvin's  Scale. 

212°,  and  the  Centigrade  scale  on  which  the  temperature  of 
melting  ice  under  standard  conditions  is  0°  and  that  of  boiling 
water  under  standard  conditions  is  100°.  Neither  of  these  start 
at  the  true  zero  of  temperature.  In  Centigrade  units  the  true 
or  absolute  zero  is  at  —  273°  C.  and  in  Fahrenheit  units  the  zero 
is  at  —  459.6°  F.  The  relative  size  of  degrees  in  the  two  systems 
is  given  by 

1°  C  =  %°  F 

It  will  be  shown  that  Kelvin's  Absolute  Scale  is  the  same  as 
that  determined  by  the  hydrogen  thermometer. 

In  looking  at  the  diagram  from  which  the  Kelvin  scale  is 
fixed  it  will  be  seen  that  the  area  beneath  any  of  the  isothermals 
and  the  adiabatics  is  equal  to  the  heat  on  the  line  and  that 
these  areas  are  equal  to  T  times  the  unit  area  e.  Hence  the 


FUNDAMENTAL  THERMODYNAMICS 


9 


heat  added  on  any  isothermal  between  points  of  intersection 
with  two  adiabatics  is  proportional  to  the  temperature  on  that 
isothermal.  This  is  an  important  consideration. 


Q  =  eT 


(14) 


is  the  expression  for  the  heat  added  on  an  isothermal  of  tem- 
perature T  between  two  adiabatics  and 

Q'  =  eT' 

is  true  at  temperature  T'  between  the  same  two  adiabatics. 
This  leads  to  the  efficiency  of  the  Carnot  cycle. 

CARNOT  CYCLE 

Carnot  proposed  a  cycle  made  up  of  isothermal  and  adiabatic 
lines  in  a  peculiar  form  of  engine.  Imagine  a  perfect  non-con- 
ducting cylinder  with  a  perfectly  conducting  head  containing  a 
non-conducting  piston  together  with  two  bodies  S  and  R  of  infinite 


FIG.  4. — Carnot  engine. 

capacity  for  heat  and  a  non-conducting  plate  P  (Fig.  4).  The 
fact  that  S  and  R  are  of  infinite  capacity  means  that  if  a  finite 
amount  of  heat  is  added  or  taken  away  from  them  their  tem- 
peratures will  not  change.  The  body  S  being  at  a  higher  tem- 
perature, Tij  than  that  of  the  body  R  at  Tz,  is  known  as  the 
source  while  R  is  called  the  refrigerator. 

If  now  the  cylinder  C  is  placed  on  the  body  S  and  the  pressure 
on  the  piston  rod  is  made  a  differential  amount  less  than  that 
exerted  by  the  substance  within  the  cylinder  on  the  piston,  the 
piston  will  be  driven  upward.  This  would  mean  that  work  is 
done  by  the  substance  within  the  cylinder  and  this  would  im- 
mediately cause  its  temperature  to  fall.  On  account  of  the 


10  HEAT  ENGINEERING 

perfect  conductor  forming  the  head  of  the  cylinder,  heat  will 
flow  at  once  into  the  substance  and  keep  its  temperature  con- 
stant, so  that  if  this  operation  is  allowed  to  go  on,  a  certain 
amount  of  external  work  would  be  done,  a  certain  amount  of 
heat  would  be  abstracted  from  S  and  the  substance  in  the 
cylinder  would  be  left  in  a  given  condition.  If  the  pressure  on 
the  piston  rod  in  this  second  condition  is  kept  at  a  differential 
amount  above  the  pressure  exerted  on  the  piston  by  the  sub- 
stance, the  piston  will  move  downward  compressing  the  sub- 
stance within  and  thus  doing  work  upon  it,  which  tends  to  in- 
crease the  temperature  causing  a  flow  of  heat  into  S  the  tem- 
perature of  which  cannot  change.  After  the  starting  point 
is  reached  the  substance  within  the  cylinder  is  in  its  original 
condition  with  a  restitution  of  the  heat  taken  from  S  and  the 
development  of  the  first  external  work  on  the  substance  within. 
These  two  actions  out  and  back  are  isothermal  and  because  (a) 
they  can  be  imagined  to  take  place  in  either  direction,  with 
(6)  external  conditions  differing  by  an  infinitesimal  quantity 
and  because  (c)  all  things  external  and  internal  can  be  restored 
to  the  initial  conditions  by  coming  back  over  the  path  to  the 
original  condition  by  a  reversal  of  directions;  such  action  is 
known  as  reversible  action.  These  three  conditions  must  hold 
for  all  reversible  actions. 

If  instead  of  returning  to  the  original  condition  after  the  addi- 
tion of  Qi  heat  units  from  S,  the  cylinder  is  slipped  to  the  non- 
conducting plate  and  then  the  substance  within  the  cylinder  is 
allowed  to  expand  by  causing  the  pressure  on  the  rod  to  be  a 
differential  amount  less  than  the  pressure  on  the  piston,  the 
substance  within  can  receive  no  heat  and  the  expansion  will  be 
adiabatic.  The  external  work  will  be  done  at  the  expense  of  or 
by  the  intrinsic  energy  in  the  substance.  Of  course  the  tem- 
perature within  the  body  must  fall  and  when  this  reaches  the 
temperature  of  R,  (T2),  the  operation  is  stopped  and  the  cylinder 
is  put  into  communication  with  R. 

If  before  the  cylinder  is  put  into  communication,  the  force 
on  the  piston  rod  is  increased  an  infinitesimal  amount,  the  sub- 
stance within  the  cylinder  will  be  compressed  by  external  energy 
applied  and  this  operation  will  be  adiabatic,  taking  place  in  the 
reverse  direction  over  the  same  path.  It  is  seen  that  this  ful- 
fills the  three  conditions  mentioned  above  and  hence  the  adiabatic 
line  is  reversible. 


FUNDAMENTAL  THERMODYNAMICS  11 

When  the  cylinder  is  connected  with  R  and  the  force  on  the  rod 
is  increased  by  a  differential  amount,  the  substance  within  is 
compressed.  This  tends  to  increase  the  temperature  which 
causes  a  flow  through  the  perfect  conductor  into  the  body  of  in- 
finite capacity.  This  causes  reversible  isothermal  compression. 
Suppose  that  this  is  continued  until  Qz  heat  units  are  abstracted 
to  such  a  point  that  when  the  cylinder  is  removed  to  P  and 
adiabatic  compression  to  the  temperature  T\  occurs,  it  will  be 
in  its  original  condition.  This  is  shown  on  the  pv  plane  by  Fig. 
5.  In  this  1-2  is  an  isother- 
mal of  temperature  Ti  on 
which  Qi  heat  units  have  been 
added.  Point  2  is  arbitrary. 
2-3  is  an  adiabatic  on  which 
no  heat  has  been  added.  The 
point  3  is  fixed  by  its  tem- 
perature Tz  at  which  the  adia- 
batic expansion  must  stop. 


3-4  is  an  isothermal  of  tern-  FlG   5._Cari^t  cycle, 

perature  Tz  on  which  Q2  heat 

units  have  been  abstracted  and  on  which  the  point  4  is  fixed  so 
that  the  adiabatic  4-1  on  which  no  heat  is  added  will  end  at 
the  original  point  1.  This  is  known  as  the  Carnot  cycle. 

REVERSED  CYCLE 

There  is  no  reason  why  these  operations  could  not  occur 
in  the  reverse  direction.  In  this  case  Q2  heat  units  would  be 
taken  from  R  and  Qi  heat  units  would  be  given  up  to  S. 

Consider  the  path  1,  2,  3,  4,  1  on  which  the  total  heat  added  is 

Qi  +  0  -  Q2  +  0  =  d  -  Q2 
but 

Q  =  Ub  -  Ua+  AW 

and  on  the  closed  path 

Ub  =  Ua 

since  the  operation  is  brought  back  to  the  original  point. 
Hence 

Qi  -Q2  =  AW 

If  this  takes  place  in  the  opposite  direction 
Q2  -  Qi  =  -  AW 


12  HEAT  ENGINEERING 

or  work  AW.  must  be  done  from  the  outside  so  that  Qi  heat 
units  may  be  discharged  into  the  source  when  a  smaller  quantity 
Q2  has  been  added  from  the  refrigerator. 

Of  course  these  must  be  so  since  if  the  substance  within  the 
cylinder  has  been  brought  back  to  its  original  condition  and  Qi 
heat  units  have  been  added  while  Q2  have  been  abstracted,  the 
difference  Qi  —  Qz  must  have  gone  into  external  work.  In 
any  case  since 

J  CdQ  =  U2-Ui  +  CdW 

J  (dQ  =  CdW  (15) 

if  C/2  —  Ui  =  0.  In  any  closed  cycle  (a  path  ending  at  the 
point  from  which  it  starts)  the  algebraic  sum  of  the  heats  on  all 
paths  of  the  cycle  is  equal  to  the  work  done.  This  is  regardless 
of  the  reversibility  of  the  path. 

The  heat  supplied  on  any  cycle  is  Qi  and  hence  its  efficiency  is 


For  the  Carnot  cycle  this  may  be  simplified,  since  the  heat 
transfers  are  on  isothermals  limited  by  the  same  two  adiabatics. 
In  this  case 

Qi  =  eTi  [See  Eq.(14)] 


This  is  the  value  of  the  efficiency  of  the  Carnot  cycle  and  it  is 
the  expression  for  availability  as  will  be  seen  later.  Hence  be- 

T"  __  rp 
tween  the  limits  TI  and  T2,  heat  has   an  availability     * 

if  used  on  a  Carnot  cycle. 

MAXIMUM  EFFICIENCY 

Suppose  there  is  a  cycle  of  efficiency  greater  than  that  of 
the  Carnot  cycle.  Call  this  cycle,  cycle  r.  If  there  is  a  source 
and  a  refrigerator  of  temperatures  TI  and  T2,  the  Carnot  cycle 
and  other  cycle  may  be  operated  between  these,  and  the  efficiencies 
will  give  the  inequality 

Qlc   -   Q*c    ^   Qlr   ~    Q2r 
Qlc  <  Olr 


FUNDAMENTAL  THERMODYNAMICS  13 

Suppose  these  two  engines  be  connected  to  the  same  shaft 
and  the  efficient  engine  be  used  as  the  driver  to  drive  the 
Carnot  engine  in  a  reversed  direction.  In  this  case  Qir  will  be 
taken  from  S  and  Q\c  will  be  given  to  the  source.  Neglecting 
friction  the  work  required  by  one  will  just  equal  the  work 
developed  by  the  other. 

do  -  Q2c  =  Qir  -  Q2r 
Since  the  inequality  is  true 

Qlc    >    Qir 

or  more  heat  is  being  added  to  the  source  than  is  taken  from  it. 
Since  the  two  engines  coupled  to  the  source  and  refrigerator  re- 
ceive no  energy  or  give  no  energy  to  the  outside  (one  drives  the 
other)  the  heat  going  to  the  source  must  come  from  the  re- 
frigerator, and  there  exists  something  with  no  connection  with 
outside  systems  which  causes  heat  to  flow  from  a  point  of  low 
temperature  to  a  point  of  higher  temperature  when  placed  be- 
tween the  two  points.  This  is  unthinkable  from  all  experience, 
hence  Qic  cannot  be  greater  than  Qir  and  the  efficiency  of  the 
cycle  cannot  be  greater  than  that  of  the  Carnot  cycle.  Hence 
the  Carnot  cycle  is  as  efficient  as  any  cycle. 

Suppose  now  the  cycle  is  reversible,  in  which  case  the  Carnot 
cycle  might  be  used  as  the  driver  if  of  greater  efficiency  than  that 
of  the  other  cycle. 

Qic    -    Q2c  Qir   -    Q2r 

Qlc  Qir 

V  Qlc    -    Q2c    =    Qir    -   Q2r 

Qir    >    Qlc 

Or  the  amount  received  by  the  source  from  the  reversed  cycle 
is  greater  than  that  taken  by  the  direct  Carnot  engine.  This 
cannot  be  so  and  hence  a  Carnot  cycle  cannot  have  a  greater 
efficiency  than  that  of  the  reversible  cycle.  But  none  can  be 
greater  than  the  Carnot. 

Since  the  efficiency  of  the  cycle  r  cannot  be  greater  and  can- 
not be  less  than  that  of  the  Carnot  cycle  it  must  be  equal  to  it 
and  hence  all  reversible  cycles  have  the  same  efficiency,  which  is 
equal  to  that  of  the  Carnot  cycle.  All  that  can  be  said  of  non- 
reversible  cycles  is  that  they  cannot  have  greater  efficiencies  than 
that  of  the  Carnot  cycle. 


14  HEAT  ENGINEERING 

This  proof  -can  be  used  to  show  that  the  efficiency  is  inde- 
pendent of  the  medium  used,  for  if  one  substance  gave  a  higher 
efficiency  it  could  be  used  on  a  Carnot  cycle  as  a  driver  while  the 
other  substance  would  be  used  as  the  medium  of  a  reversed 
Carnot  cycle.  The  reasoning  would  lead  to  the  same  absurd 
result  and  hence  all  substances  will  give  the  same  efficiency  theo- 
retically when  operating  between  the  same  source  and  refrigerator. 

The  Carnot  efficiency  is  the  maximum  efficiency  and  if  the 
limiting  temperatures  are  Ti  and  T2 


is  the  maximum  availability.  Being  the  maximum  it  is  that 
which  should  be  obtained  in  practice  theoretically  and  hence 
it  is  called  simply  the  availability. 

CONDITIONS  FOR  AVAILABILITY  OF  HEAT 

This  leads  to  the  observation  that  the  only  way  in  which  heat 
can  be  available  is  to  have  a  difference  in  temperature  necessi- 
tating two  bodies  at  different  temperatures  and  a  substance 
which  can  receive  and  discharge  the  heat.  The  three  systems 
are  necessary  to  make  heat  available. 

(m    _   rji  \ 
-^f,  —  -  JQi  are  available  for  useful  work 

m 

and    QIY  must  be  rejected.     Of  course  this  rejected  heat  or 

wasted  heat  so  far  as  the  work  ATT  is  concerned  may  be  used 
for  some  valuable  purpose,  as  when  it  is  rejected  from  a  steam 
engine  and  used  for  warming  buildings  or  for  heating  water  for 
a  wash  house. 

This  leads  to  the  Second  Law  of  Thermodynamics: 
SECOND  LAW 

"It  is  impossible  by  means  of  a  self-acting  machine  unaided  by  any 
external  agency  to  convey  heat  from  one  body  to  another  at  higher 
temperature"  or  "no  change  in  a  system  of  bodies  that  can  take  place 
of  itself  can  increase  the  available  energy  of  the  system."  When  the 
matter  is  discussed  in  the  manner  given  above,  this  law  results  in  the 
expression 

m     _   rp 

—  ^  —  =  Availability  or  Carnot  Efficiency, 

,„     dQ       dH 
dS  =  -r  +  -r 


FUNDAMENTAL  THERMODYNAMICS  15 

is  the  conclusion  from  this  law  when  the  gain  in  unavailable  energy  is 
considered,  as  will  be  seen  later.  Each  of  the  above  expressions 
results  from  the  second  law  of  thermodynamics. 

Having  seen  two  paths  which  are  reversible,  the  isothermal 
and  the  adiabatic,  it  may  be  well  to  consider  some  changes  which 
are  irreversible  and  for  that  reason  cannot  truly  be  represented 
on  the  pv  plane. 

LOSS  OF  AVAILABILITY 

Suppose  heat  flows  by  conduction  from  one  body  of  high  tem- 
perature to  one  of  low  temperature.  Such  action  cannot  possibly 
be  assumed  to  take  place  in  the  reversed  direction  and  therefore 
on  account  of  the  first  requirement  it  is  non-reversible. 

If  a  gas  is  assumed  to  discharge  freely  from  a  vessel  of  pressure 
pi  to  a  place  of  distinctly  lower  pressure  p%,  such  action  could  not 
be  assumed  to  take  place  in  an  opposite  manner,  and  lastly  if 
work  be  changed  into  heat  as  by  friction  this  could  not  be  as- 
sumed to  take  place  in  a  reverse  direction. 

In  all  of  these  there  is  a  loss  of  availability. 

In  the  first  the  loss  of  available  heat  would  be  the  difference 
of  the  available  heats  at  the  two  temperatures.  This  is 


av 

if  To  is  the  lowest  possible  temperature,  and  T\  and  Tz  are  the 
temperatures  of  the  two  sides  of  the  conducting  surface. 

In  the  second  case  the  pressure  has  been  brought  nearer  to 
the  pressure  of  the  surrounding  medium  and  hence  by  the  time 
it  is  brought  to  this  pressure  by  adiabatic  expansion  the  tem- 
perature will  not  be  so  low  as  it  could  have  been  before.  Hence 


4  -  9]  >  4  - 


Since 

To  <  To' 
Ti  is  almost  equal  to  TV 

Loss  in  available  heat  =  TQ  jf, To'  ^r 

J-l  1\ 

In  the  third  case  if  ATT  units  of  work  are  changed  into  heat 
of  temperature  T\  the  available  part  of  this  is 


16  HEAT  ENGINEERING 

All  of  it  was  available  as  mechanical  energy  before  the  change, 
hence 


represents  the  loss  of  available  energy. 

In  the  reversible  changes  or  on  the  reversible  cycles  the  un- 
available energy  before  the  operation  was 

a  Ta 

Q.jT 

and  this  is  the  amount  rejected  by  the  reversible  cycle.     Hence 
in  this  case  there  is  no  increase  of  unavailable  energy. 

ENTROPY 

Goodenough  points  out  that  in  all  of  these  cases  the  non- 
reversible  changes  have  brought  about  a  loss  of  available  energy 
and  that  these  expressions  are  all  of  the  form 


He  then  says  that  the  quantity  which  when  multiplied  by  T0, 
the  lowest  available  temperature,  gives  the  increase  of  un- 
available energy  due  to  a  non-reversible  or  other  change  is  called 
the  increase  of  entropy. 

This  quantity  is  of  the  form  ^  and  refers  to  one  system  or  to 

several  systems.  If  one  system  alone  is  considered  there  may 
be  an  increase  or  decrease  of  entropy  due  to  the  change  of 
heat,  while  if  two  systems  are  considered  reversible  changes  lead 
to  no  change  in  unavailable  energy  and  hence  no  change  of 
entropy,  while  non-reversible  changes  lead  to  an  increase  of  un- 
available energy  and  consequently  an  increase  of  entropy. 
Thus  if  the  source  alone  is  considered  when  Qi  heat  units  are 
given  up  to  the  cylinder  of  the  Carnot  engine,  there  is  an  un- 
available amount  of  heat  T0  >~  taken  from  this  source  and 

1 1 

given  to  the  medium  in  the  cylinder.     There  is  a  decrease  of 

entropy  of  w^  for  the  source  but  that  of  the  medium  is  increased 
1 1 

by  ~-  and  the  sum  of  these  two  is  zero.     This  is  true  if  the  cycle 
i  i 


FUNDAMENTAL  THERMODYNAMICS  17 

is  worked  in  the  reverse  direction.     If  there  is  conduction  of 

Oi 


heat  the  loss  of  entropy  will  be  ~,  while  the  gain  of  entropy  will 
be  7^-.     This  latter  is  gr 

-I  2 

there  is  a  gain  in  entropy 


be  7^-.     This  latter  is  greater  since  T2  is  less  than  TI.     Hence 

-I  2 


when  the  irreversible  change  takes  place.  This  is  the  only 
direction  in  which  this  operation  can  take  place,  while  in  re- 
versible action  the  operation  may  take  place  in  either  direction. 
Hence  if  all  bodies  or  systems  taking  place  in  a  change  be  con- 
sidered together,  a  process  which  will  take  place  of  itself  will  be 
accompanied  by  an  increase  of  entropy.  If  there  is  no  change 
in  the  entropy  the  change  will  take  place  in  either  direction. 

Goodenough  shows  further  that  only  one  part  of  a  system 
may  be  considered  and  then  the  increase  of  unavailable  energy 


Q 


fdQ 

°J    T 


may  be  accomplished  by  adding  heat  Q  or  if  the  temperature 
TI  is  not  constant  by  adding  I  dQ.  Since  this  heat  may  come 

from  internal  friction,  as  well  as  from  the  outside,  and  the 
symbol  Q  refers  to  heat  added,  H  will  be  used  for  the  heat  de- 
veloped by  internal  friction.  In  this  case  then 

CdQ  ,     CdH 
Entropy  change  =    I  -^  +    I  -^ 

There  could  be  a  loss  of  availability  and  consequently  an  in- 
crease of  entropy  if  there  was  internal  conduction,  but  this  need 
not  be  considered  as  in  most  problems  the  system  is  assumed  at 
a  uniform  temperature  throughout  and  any  change  in  one  part 
is  the  same  in  all  parts.  If  S  be  assumed  for  the  symbol  of 
entropy 


Since  the  amount  of  unavailable  energy  is  dependent  on  the 
energy  in  a  body  and  its  temperature,  the  entropy,  which  is 
the  unavailable  energy  divided  by  the  lowest  possible  tern- 


18  HEAT  ENGINEERING 

perature,  is  also  dependent  on  these.  Hence  entropy  depends 
on  the  state  of  a  body  and  dS  must  be  an  exact  differential, 
giving 


Ti  UTi 

ENTROPY  AROUND  A  CYCLE 

Now  dH  is  the  heat  of  internal  friction  and  is  positive  in  all 
cases.     Hence  although  around  a  closed  cycle    I  dS  =  0,  since  a 

return  is  made  to  the  original  point,     I  -^-   must  always  be 
positive  and  therefore  in  such  a  case       -~~  must  be  a  negative 


CdQ  . 

\  -^    is  negatr 


quantity.     When  there  is  internal  friction     I  -^    is  negative 

around  a  closed  cycle.  If,  however,  dH  =  0  there  is  no  friction 
and 

'f  =  0  (19) 

around  a  closed  cycle.  This  friction  when  it  exists  always  acts 
to  produce  heat  no  matter  in  what  direction  one  progresses  along 
a  path.  It  is  this  which  makes  a  theoretical  path  of  any  form 
on  the  pv  plane  an  irreversible  path.  Hence  the  following 
statements  are  made: 

on  any  closed  cycle. 

^=0 


C 

=    I 


on  a  closed  reversible  cycle. 

S*  -  Si  =  0  >  I  ^  (20) 


on  a  closed  irreversible  cycle  (internal  friction  present). 
Now 

dS  =  ^  +  — 
or 

r          r         f 

dH 


J7US  =  idO   +  I 


FUNDAMENTAL  THERMODYNAMICS  19 

ENTROPY  DIAGRAM 

TdS  is  an  area  between  a  curve  of  coordinates  T  and  S 

and  the  axis  of  entropy  and  it  must  equal  the  heat  added  from 
the  outside  if  dH  =  0  or  the  line  is  reversible  and  if  dH  does 
not  equal  zero  it  represents  the  heat  added  from  the  outside, 
plus  the  heat  added  by  friction. 

With  these  coordinates,  lines  of  constant  temperature  become 
parallel  to  the  S  axis  or  horizontal  while  lines  of  constant  entropy 
are  vertical  lines.  The  areas  beneath  any  line  represent  the 
heat  added  from  the  outside  plus  the  internal  friction.  If 
there  is  no  friction  (the  line  is  reversible),  area  beneath  the  line 
represents  the  heat  added  from  the  outside.  If  the  lines  are 
reversible  the  line  of  constant  entropy  must  be  an  adiabatic  since 

'dQ 


and  because  T  does  not  equal  infinity,  dQ  must  equal  zero.  This 
is  the  condition  for  an  adiabatic.  If  the  lines  are  adiabatic  lines 
with  internal  friction  the  entropy  increases  due  to  this  friction 
and  hence  the  lines  must  progress  to  the  left. 

CHARACTERISTIC  EQUATIONS 

A  substance  of  unit  weight  under  a  definite  hydrostatic  pressure 
and  at  a  given  temperature  will  occupy  a  certain  volume  and  if 
the  pressure  and  temperature  change,  the  volume  will  change. 
In  most  cases  there  is  for  every  substance  a  functional  relation 
between  these  three  properties  or 

0(p,  v,  t,)  =  0  (21) 

This  is  called  the  characteristic  equation  of  the  substance 
and  for  each  substance  there  is  assumed  to  be  such  an  equa- 
tion determined  by  experiment.  This  equation  being  between 
three  coordinates  is  the  equation  of  a  surface. 

The  equations  for  some  of  the  well-known  substances  used  in 
heat  engines  are  as  follows: 
For  perfect  gases 

pv  =  BT  or  (22) 

pV  =  MET  (23) 


20  HEAT  ENGINEERING 

For  saturated  steam 

T 

log  p  =  log  k  -  n  log  T  _  ^  (24) 

For  saturated  vapors 

log  p  =  a  -f  ban  +  c@n  (25) 

where  n  =  (t  —  t0) 

For  superheated  steam 


v  +  c  =  y-    -  (1  +  ap)  (26) 

For  other  superheated  vapors 

pv  =  BT  -  cpn  (27) 

HEAT  ON  A  PATH 

Now  to  study  any  path  on  which  heat  is  added  the  path  on 
the  surface  represented  by  the  characteristic  equation  could  be 

projected  on  any  one  of  the  three 
planes,  pv,  pt  or  vt.  The  first  one  of 
these  is  the  best  since  area  beneath 
the  path  represents  external  work  but 
further  than  this  there  is  no  reason  for 
its  use.  Suppose  for  instance  the  path 
is  shown  in  Fig.  6.  To  find  the  heat  the 


v  path  is  changed  to  the  broken  path  of 

IG.    .     p^aneon      e  p         constant  pressure  and  constant  volume 

lines  of  differential  length  as  shown. 
The  heat  to  change  the  volume  by  unit  amount  on  the  constant 

pressure  line  is  (— )  and  the  amount  to  change  the  pressure  by 

\   OV  I     77 


unity  on  a  constant  volume  line  is  (-7— )  .     In  the  broken  path 

\0p/  v 

the  change  from  a  to  1  is  dv  and  that  from  1  to  6  is  dp.  These 
quantities  are  arbitrary  in  amount  if  the  path  is  not  fixed  and 
hence  they  are  independent  variables.  The  heat  added  on  a 
path  is 

*-(§),*+©*  <28> 

It  must  be  remembered  that  for  a  definite  path  only  one  of  these 
is  an  independent  variable  since  the  path  necessitates  a  relation 
between  the  two  variables. 


FUNDAMENTAL  THERMODYNAMICS  21 

In  the  same  manner,  if  the  path  is  projected  on  the  plane  pt, 
or  vt,  the  expressions  would  be 


or 

If  a  pound  of  substance  is  referred  to,  these  may  be  written 
with  small  letters. 

The  equations  are  the  fundamental  differential  equations 
of  heat. 


The  quantities  (TT)  ,   etc.,   which  represent  the  amount   of 

heat  added  to  1  Ib.  of  substance  to  change  one  of  the  proper- 
ties, p,  v,  or  t,  by  unity  when  another  property,  p,  v,  or  t,  remains 
constant,  are  known  as  thermal  capacities.  Since  these  are  ex- 
pressed in  heat  units  which  refer  to  water  their  values  have  the 
definite  names  given  below 

f  ~~J   =  cv,  specific  heat  at  constant  volume. 

— )    =  cp,  specific  heat  at  constant  pressure. 

ot  I  p 

These  are  the  amounts  of  heat  to  change  the  temperature 
of  unit  weight  by  one  degree  when  either  the  volume  or  pressure 
is  constant. 


~ )  =  lv,  latent  heat  of  expansion. 

OV/  i 

This  is  the  amount  of  heat  added  to  1  Ib.  of  substance  at 
constant  temperature  to  change  the  volume  by  unity.  It  is 
latent  because  the  temperature  does  not  change. 

\~~ }  =  lp,  latent  heat  of  pressure  change. 

(—  j    =  n,  heat  of  expansion  at  constant  pressure. 

I-—-]   =  o,  heat  of  pressure  change  at  constant  volume. 

Using  these  symbols  the  three  equations  may  be  simplified 

dq  =  cvdt  +  lvdv  (290 

dq  =  Cpdt  +  Ipdp  (300 

dq  =  ndv  +  odp  (280 


22  HEAT  ENGINEERING 

It  will  be  remembered  that  dq  is  not  an  exact  differential 
and  hence  there  is  no  differential  relation  between  the  coefficients 
of  any  one  of  the  equations.  There  are  relations  which  may  be 
made  by  equating  any  two  of  the  equations  since  these  are  ex- 
pressions for  the  same  quantity.  Hence 

cvdt  +  lvdv  =  cpdt  +  Ijdp  (31) 


This  appears  to  be  an  equation  with  three  independent  vari- 
ables. There  is  always  a  characteristic  equation  with  p,  v,  and 
t,  and  hence  any  one  of  these  three  is  known  in  terms  of  the 
other  two.  Suppose  t  is  eliminated  by 


Hence 

r  /dt\  ,  7~ij       /8t\       r  /8t\ 

cJ  — )   +  lv  \dv  +  cv(  — )  dp  =    cJ  — )  + 
L     \ozv  p  \0£>/ 1-          L     \dp/  „ 

Now  this  equation  is  true  for  any  value  of  dv  and  for  any  value 
of  dp  if  a  definite  path  is  not  assumed,  or  in  other  words,  it  is  true 
for  any  path.  The  only  way  that  this  can  be  true  is  to  have 
the  coefficients  of  the  same  variable  on  each  side  of  the  equation 
equal.  Hence 


or 

cp  —  cv  =  —  M|?)  (32) 

and 


or 

7      II  /OO  \ 

cp  —  cv  =  MTT)  (33) 

\ol/  p 

Eliminating  dv  from  (31)  leads  to 


(34) 

and 

/A»I\ 

(33) 


FUNDAMENTAL  THERMODYNAMICS  23 

Eliminating  dp  from  (31)  leads  to 


and 


which  have  been  found  before.     By  equating  (28')  with  either 
(29;)  or  (30')  other  relations  could  be  found. 

(35) 
(36) 


(St\ 

0    =   CJ  — ) 

\8pJ  v 


DIFFERENTIAL  EQUATIONS  AND  RELATIONS 

Now 

Jdq  =  du  +  pdv 

and  for  reversible  paths  (dH  =  0) 

dq  =  Tds 

Tds 
.'.  du  =  Jdq  —  pdv  = pdv  (37) 

A. 

Another  quantity  used  throughout  the  theory  of  thermody- 
namics which  has  an  important  use  is  the  heat  content.  This  is 
not  the  heat  contained  in  a  body  but  it  is  that  heat  plus  the 
product  of  the  pressure  and  volume.  This  is  the  amount  of 
energy  which  would  leave  with  a  substance  when  it  is  forced 
out  of  a  region  in  which  the  pressure  is  kept  constant.  It  is 
represented  by  the  symbol  i  and  denned  by 

i  =  A(u  +  pv)  (38) 

/  =  Mi 
.  M  =  weight  of  the  substance. 

It  will  be  seen  that  i  depends  on  the  state  of  the  substance 
since  u,  p}  and  v  are  all  dependent  on  the  state.  It  is  expressed 
in  heat  units. 

di  =  Adu  +  Apdv  +  Avdp 

Substituting  from  (37) 

di  =  Tds  +  Avdp  (39) 


24  HEAT  ENGINEERING 

Two  other  quantities  known  as  thermodynamic  potentials, 
F  and  $,  are  given  by  the  equations 

F  =  Au  -  Ts 
<t>  =  Ai  -  Ts 

These  are  both  functions  of  the  state,  since  u,  T,  s,  and  i  are 
fixed  by  the  state.  Their  differentials  reduce  to 

-  dF  =  sdT  +  Apdv  (40) 

d<f>  =  Avdp  -  sdT  (41) 

Now  du,  di,  dF  and  d<j>  are  all  exact  differentials  since  they 
depend  on  the  state  and  hence,  as  they  are  expressed  as  functions 
of  two  variables,  the  partials  of  the  coefficients  of  the  independent 
variables  with  regard  to  the  other  independent  variables  are 
equal.  Taking  the  equations  (37),  (39),  (40)  and  (41)  the  follow- 
ing results  are  true: 

(£).-'©.  <« 

•         '«' 


The    four    equations    above    are    Maxwell    thermodynamic 
equations. 
Now 

dq 

Y  =  ds 

Hence  (44)  and  (45)  reduce  to 

*•  (46) 


Substituting  (46)  and  (47)  equations  (29')  and  (30')  become 

dq  =  cvdt  +  AT          dv  (48) 


dq  -  Cj4t  -  AT  dp  (49) 

ot/  p 


FUNDAMENTAL  THERMODYNAMICS  25 

Substituting  (36),  (35),  and  (33)  in  (28')  this  becomes 

-»>(!)  >  +  *(§.*• 

(50) 


Now 

Cr      .  -    7     ||  /'QQ\ 

P      Cv  —  Lv  I —  I  (^cJo; 

and 


and 

/M\ 

\SfV 


p  Cp  —   Cv 

u 


iu\ 

\8pJ  v        Cp  —  cv 
Equation  (50)  becomes 


(48),  (49)  and  (52)  are  in  the  same  form  since  the  partial  de- 
rivatives are  all  derivatives  with  regard  to  dT. 

Equation  (51)  is  an  important  relation  which  leads  to  valuable 
results. 

Goodenough  derives  two  other  equations  which  are  of  value. 
These  follow: 

du  =  Jdq  —  pdv 
using  (48)  this  becomes 

du  =  Jcvdt  +  [^(^)   -  p]dv  (53) 

di  =    Tds  +  Avdp 
=       dq  +  Avdp 

using  (49)  this  reduces  to 

di  =  cpdt  -  A  [T(f})    -v]dp  (54) 


26  HEAT  ENGINEERING 

du  and  di  are  exact  differentials.     Hence 


/5cv\        dp        „  5zp       dp 
\dv/t~  M  dt2  "  dt 

and  from  (54) 

(!'),-  -«©,     :'.;'•    <»' 

These  equations  (55)  and  (56)  are  of  value  in  making  certain 
deductions. 

Equations  (46),  (47),  (48),  (49),  (50),  (51),  (52),  (53),  (54), 
(55)  and  (56)  are  important  equations  in  the  development  of 
thermodynamics.  It  must  be  remembered  that  all  of  these 
are  general  equations  and  refer  to  any  substance.  The  partial 

derivatives,  such  as   I— }    ,  are  found  from  the  characteristic 
equation  of  the  substance  used  in  any  problem. 

PERFECT  GASES       ; 

A  perfect  gas  is  a  substance  which  obeys  the  law  of  Boyle  and 
the  law  of  Charles.  From  these  two  laws  the  characteristic 
equation  becomes 

pv  =  BT  (57) 

or 

pV  =  MET  (58) 

p  =  pressure  in  pounds  per  square  foot 
v  =  volume  of  1  Ib.  of  substance  in  cubic  feet 
V  =  total  volume  of  M  pounds  in  cubic  feet 
M  =  weight  in  pounds 
B  =  a  constant 
T  =  absolute  temperature 
=  Fahr.  temp.  +  459.6 

Now 


m  =  weight  of  1  cu.  ft.  of  gas. 


FUNDAMENTAL  THERMODYNAMICS  27 

If  for  air,  the  weight  of  1  cu.  ft.  is  found  to  be  0.08071  Ibs. 
per  cubic  foot  at  atmospheric  pressure  and  at  32°  F.,  B  is  found 
to  be 

2116.3 


0.08071  X  491.6 


By  the  law  of  Avogadro  equal  volumes  of  gases  at  the  same 
pressure  and  temperature  contain  equal  number  of  molecules 
hence,  if  wt  =  molecular  weight 

m  =  Kwt 
-    D  _        Po        _  const. 

=  Kwt  TQ   ~~       wt 
or 

Bwt  =  const.  =  R  (60) 

R  =  universal  gas  constant. 

The  weight  of  oxygen  per  cubic  foot  under  atmospheric 
pressure  and  at  32°  F.  is  0.089222  Ibs. 

_  2116.3    __ 

B  oxygen     ~    Q  Q89222   X   491.6 

Hence 

R  =  48.25  X  32  =  1544 
and 


Now  the  molecular  weight  of  air  is  found  from  the  fact  that 
air  contains  79  per  cent,  by  volume  of  N%  and  21  per  cent,  of 
02.  By  Avogadro's  Law,  considering  one  molecule  to  each  unit 
volume,  the  weight  is  equal  to  volume  multiplied  by  the  molecular 
weight. 

0.79  X  28.08    =  22.18 

0.21  X  32         =    6.72 


wt.  of  1  vol.  =  28.90 
Hence 

R       -  15*4  _ 

#«*>  ~  28.9  " 

The  correct  value  of  this  is  53.34  showing  that  the  molecular 
weight  of  air  is  28.93. 

If  there  is  a  mixture  of  various  gases  in  a  given  volume  there 
are  two  points  of  view  to  be  considered:  (a)  all  of  the  constituent 


28  HEAT  ENGINEERING 

gases  occupy  the  whole  volume  and  each  is  under  a  partial 
pressure  such  that  f  the  sum  of  the  pressures  equals  the  total 
pressure,  or  (6)  each  gas  is  under  the  total  pressure  but  occupies 
a  part  of  the  volume  so  that  the  sum  of  the  partial  volumes 
equals  the  total  volume.  The  first  view  is  known  as  Dalton's 
Law  and  represents  what  really  takes  place  when  a  number  of 
gases  are  mechanically  mixed.  If  pi,  p%,  ps  .  .  .  represent  the 
partial  pressures  of  the  various  gases  which  weigh  MI,  M2,  M3 
.  .  .  and  if  all  the  gases  occupy  the  volume  V  the  following 
are  true: 

p  =  PI  +  p%  +  PS  .  .  •  (Dalton's  Law) 
M  =  Mi  +  M2  +  M3  .  .  . 

p  =  M1B1^  + 

V 

Pp  =  MBmixture  = 

or 

R  2MlBl 

^mixture    ~ 


This  is  not  as  convenient  a  method  as  working  with  volumes 
since  in  most  cases  it  is  the  proportional  volumes  which  are 
known  and  not  the  proportional  weights.  By  Avogadro's  Law 
each  unit  of  volume  may  be  assumed  to  have  one  molecule. 

proportional  weight. 

2  proportional  weights  =  total  weight 

If  this  is  divided  by  the  sum  of  the  volumes,  the  weight  of 
unit  volume,  which  by  assumption  is  the  molecular  weight,  is 
found. 
Hence 

ZViwh  . 

J\V          ~   Wlmixture  WW 

I544 


&  mixture   ~ 

In  the  above  manner  by  (61),  (63)  and  (64)  the  value  of  B 
for  any  gas  or  mixture  may  be  found  if  the  gas  is  assumed  to  be  a 
perfect  gas.  No  gas  obeys  the  laws  of  Boyle  and  Charles  com- 
pletely and  hence  the  characteristic  equation 

pV  =  MET 

is  not  exactly  true  for  any  gas.     Under  great  pressures  there  is 
variation,  but  air,  hydrogen,  nitrogen,  carbon  monoxide,  meth- 


FUNDAMENTAL  THERMODYNAMICS  29 

ane,  ethylene,  and  other  hydrocarbons  are  assumed  to  follow 
these  laws.  Carbon  dioxide  and  steam  do  not  obey  the  equa- 
tion, but  at  times  they  are  handled  as  if  they  did. 

To  aid  in  working  out  the  values  of  B  the  following  molecular 
weights  are  given: 

Hydrogen  H2  2.016 

Carbon  monoxide  CO  28.0 

Oxygen  O2  32.0 

Methane  CH4  16.032 

Ethylene  C2H4  28.032 

Nitrogen  N2  28.08 

Carbon  C  12.0 

Ammonia  NH3  17.064 

Carbon  dioxide  CO2  44.0 

Steam  H2O  18.016 

Sulphur  S  32.06 

DIFFERENTIAL  HEAT  EQUATIONS  FOR  PERFECT  GASES 

For  a  perfect  gas 

pv  =  MET 
or 

pv  =  BT 
and  the  various  equations  (46),  etc.,  are  reduced  by  the  following: 


Sv\        B 


It  is  to  be  remembered  that  although  t  +  459.6  =  T 
U  =  5T. 


ATB 

dq  =  cvdt  +  '  -  dv  =  cvdt  +  Apdv  (48) 

A  TB 

dq  =  cpdt  -   ^—dp  =  cpdt  -  Avdp  (49) 

dq  =  cp^dv+cv^dp  =  cpT^  +  cvT^  (50) 

cp-cv  =  AT~       =  AB  (51) 


30  HE  A  T  ENGINEERING 

In  equation  (48)  the  last  term  represents  the  external  work, 
therefore  the  first  term  represents  the  change  in  intrinsic  energy  : 

Jcvdt  =  du  (65) 

Now  du  is  an  exact  differential  and  must  be  directly  integrable. 
Hence  cvdt  must  also  be  directly  integrable  (i.e.,  with  no  reference 
to  any  path)  and  therefrom  cv  for  a  perfect  gas  is  a  constant 
or  a  function  of  t. 

SPECIFIC  HEATS 
Now  experiment  proves  that  cv  is  a  function  of  t  of  the  form 

cv  =  a  +  bt 
or 

cv  =  a  +  b(T  -  459.6)  =  a  -  459.66  +  bT 

=  a'  +  bT  (66) 

But 

cp  —  cv  =  AB 

cp  =  AB  +  a'  +  bT  =  a"  +  bT  (67) 

Now  since 

cp  —  cv  =  AB 

cp,  cv,  and  AB  must  be  of  the  same  nature,  if  cp  and  cv  are 
specific  quantities  AB  must  be  a  specific  quantity.  Since  cp 
is  the  amount  of  heat  to  change  the  temperature  one  degree  and 
since  cvdt  —  du  and  therefore  cv  is  the  amount  of  internal  energy 
change  when  the  temperature  changes  one  degree  on  any  line, 
AB  must  be  the  amount  of  external  work  when  the  temperature 
changes  one  degree  at  constant  pressure. 
Since 

wt  X  B  =  R,  a  constant 

wt  X  cv  =  ain  +  blu  T 

wt  X  cp  =  a™   +  b™T 

are  the  universal  forms  of  specific  heats  and  these  differ  by  AR. 


1544 
wt  X  c    -  wt  X  cv  =  AR  =  -         =  1.9855 


The  values  of  the  a's  and  6's  are  given  below 
wt  X  cv  =  4.77  +  0.000667* 
wt  X  cp  =  6.75  +  0.000667* 

For  C02 

cv  =  0.15    +  0.000066* 
cv  =  0.195  +  0.000066* 


i 

f  for  any  perfect  gas. 


FUNDAMENTAL  THERMODYNAMICS  31 

For  superheated  steam.     (Approximate.) 

cv  =  0.324  +  0.000133* 
cp  =  0.435  +  0.000133* 
Although 

cp  —  cv  =  AB 

is  always  true  the  relation 

Cf  =  k  =  1.4  (68) 

cv 

is  approximately  true  for  ordinary  temperatures.  Equations 
(51)  and  (68)  are  not  both  possible  for  two  such  equations  could 
only  hold  for  definite  values  of  cp  and  cv.  (68)  is  approximately 
true  since  the  value  of  the  quantity  b  is  small  and  as  it  gives 
simpler  results  to  certain  problems  its  use  is  advisable  unless 
there  is  a  great  change  in  temperature  on  the  path  considered. 
Since 

cp  —  cv  —  AB 
and  cp  =  kcv 

AB  =  (k-  IK 
or  cp  :  cv :  AB  =  k  :  I  :  k  —  1 

ISOTHERMAL  AND  ISODYNAMIC  LINES 

If  in  equation  (65),  T  is  made  constant  du  will  be  equal  to 
zero,  or  u  is  constant.     Hence  for  perfect  gases  the  isothermal 
is  the  same  as  the  isodynamic, 
Since 

pv  =  BT 

pv  =  constant 

is  the  equation  of  this  line  on  the  pv  plane.     This  is  the  equation 
of  the  rectangular  hyperbola. 

ADIABATIC 

The  adiabatic  is  such  a  line  that  dq  —  0. 
From  (48). 

0  =  cvdt  +  Apdv 

D/77 

cvdt  =  —  Apdv  =  —  A  -  -  dv 
cv   dt  dv 


32  HEAT  ENGINEERING 

or 


Vi\4* 


but 
hence 


;- 1 


or 

Tvk~1  =  const. 

This  is  the  equation  of  the  adiabatic  on  the  v  T  plane.     To 
reduce  this  to  the  equation  for  the  pv  plane,  T  must  be  eliminated. 
Now 

T  -PL 
~  B 

Hence 


const. 
or 


pvk  =  const.  (70) 

This  is  the  equation  of  the  adiabatic  of  a  perfect  gas  on  the 
pv  plane. 
Now 


p 

Substituting  this  in  (70)  gives 

(BT\k 
P\p)    ''=  const' 

k-l 

p  k  T  =  const. 

(69),  (70)  and  (71)  are  the  projections  of  the  adiabatic  line 
of  the  surface  pv  =  BT  (perfect  gas)  on  the  three  planes  of 
projection. 

The  line  pvn  =  const,  is  known  as  a  polytropic  and  for  the  per- 
fect gas  this  has  the  three  forms  on  the  different  planes: 

pvn  =  const.  (72) 

n-l  ,~QN 

p  »  T  =  const. 
vn~lT  =  const.  (74) 


FUNDAMENTAL  THERMODYNAMICS 


33 


THERMODYNAMIC  LINES 

The  six  important  lines  used  in  thermodynamics  are  the 
isothermal  (constant  T) ,  adiabatic  (no  heat  added  from  the  out- 
side), isodynamic  (constant  intrinsic  energy),  constant  pressure, 
constant  volume  and  polytropic. 

For  a  perfect  gas  these  lines  on  the  pv  plane  have  the  following 
equations: 

Isothermal,  pv  =  constant 

Adiabatic,  pvk  =  constant 

Isodynamic,  pv  =  constant 

Constant  pressure,  p  =  constant 

Constant  volume,  v  =  constant 

Polytropic,  pvn  =  constant 


FIG.  7. — Thermal  lines. 

It  is  seen  that  each  is  of  the  form  pvn  =  constant  in  which 
n  has  certain  values. 

n  =  1  for  isothermal  and  isodynamic 

n  =  k  for  adiabatic 

n  =  0  for  constant  pressure 

n  =  oo  for  constant  volume 

These  are  shown  in  Fig.  7. 

Since  these  are  all  of  the  same  form  it  will  be  well  to  investigate 
the  general  case  first. 


34  HEAT  ENGINEERING 

pvn  =  const. 
dq  =  cvdt  +  Apdv 
du  =  Jcvdt 

dW  =  pdv 
AB 


C    ~ 


k  - 


=  Jcv(Tz  -  TO   =   ~~  (Tz  - 


f  f»  f»  ,1-n    I" 

I  dw  =    I  pdy  =  const.     I  v~ndv  =  const.  ^ 

t/  t/ti  e/0i  -^  Bi 


0.  const. 

Since  »  =  — 

vn 

Now  Const.  =  pat)in 


1  —  n  1  —  n 

When  _  n  =  1 

Pzvz  =  piVi 

and  the  expression  for  work  reduces  to 

Work  =  g- 

an  indeterminate  expression.  To  find  the  value  of  the  work  in 
this  case,  the  integration  must  be  made  by  using  the  substitution 
for  this  case. 

Work  =    I  pdv  =  const.  I       —  =  const.  loge  —  =  piVi\oge  —  (76) 

since  const. 

P  —  ~ 

v 

The  value  (75)  for  work  holds  for  all  values  of  n  except  for 
n  =  1.  In  this  case  (76)  is  the  expression  for  work.  Equation 
(75)  may  be  put  in  two  other  forms,  thus: 


—  n  —  n 

Pzvz 


(770 


FUNDAMENTAL  THERMODYNAMICS  35 

Since 

Ml  =  Pi  y  /Pi\  --r 

or 


(78) 


__     k  —  1  1  —  n  ' 

for  all  values  of  n  except  n  =  1,  and 

q  =  p&i  loge  (80) 


The  expressions  for  the  heat  become: 


for  n  =  1  since  there  is  no  change  in  intrinsic  energy  on  the 
isothermal.     If  these  do  not  refer  to  1  Ib.  but  to  M  Ibs. 


_Iog.  (80') 

It  will  be  seen  from  (79')  that 

+  ^  --  is  proportional  to  the  heat. 
— 


—  i       i  —  n 

_  i  is  proportional  to  the  change  of  energy,  and 

is  proportional  to  the  work 


IV 


along  the  polytropic  pvn  =  const,  of  a  perfect  gas.  It  must  be 
remembered  that  this  is  only  true  for  perfect  gases  and  on 
polytropics. 

For  n  =  k,  the  expression  (79)  becomes 


_ 


k-l  l-k 


or  the  intrinsic  energy  change  *-—  I  -  J-^  is  equal  to  minus  the 

fC    -    1 

work.     If  the  final  volume  is  infinity  this  in  the  form  of  (78) 
reduces  to 


36 


HEAT  ENGINEERING 


This  is  the  work  under  an  adiabatic  to  infinity  from  the  point 
1  and  hence  it  is  the  expression  for  the  intrinsic  energy  at  the 
point  1. 

The  expressions  (75)  and  (76)  for  work  are  expressions  for 
areas  beneath  lines  and  consequently  the  expressions  are  true 
for  all  substances  expanding  on  these  paths. 

Since  p^Vz  =  piVi  on  the  polytropic  n  =  1  there  is  no  change 
in  intrinsic  energy  on  this  line  for  a  perfect  gas  and  it  is  the 
division  line  between  positive  and  negative  energy  changes. 

On  the  line  n  =  k  there  is  no  heat  added  and  so  this  is  the 
division  line  between  positive  and  negative  heat. 


-AW 


-AC 


FIG.  8. — Lines  of  division  between  positive  and  negative  quantities. 

The  constant  volume  line  is  the  division  line  between  positive 
and  negative  work. 

These  are  shown  in  Fig.  8. 


EQUALITY  OF  GAS  SCALE  AND  KELVIN  SCALE 


Now 


pv  =  BT 
has  been  derived  from  the  definition  of  the  perfect  gas  from  which 


(81) 


dt        v       T 
From  (53)  which  has  been  derived  from  absolute  considerations 

du  =  Jcvdt  +  \T  -IT  —  pldv 
L       ot 


FUNDAMENTAL  THERMODYNAMICS  37 

Since  du  is  independent  of  v  for  a  perfect  gas  by  the  experi- 
ments of  Joule  and  Thompson: 


(Joules'  Law)  T  '  -  p  =  0 

or 


(81)  and  (82)  give  the  same  result,  hence  the  absolute  Kelvin 
T  of  (82)  must  be  the  same  as  the  perfect  gas  T  of  (81). 


HEAT  CONTENT  OF  GASES 


a-  PV 

Since  u  = 


k-  1 


7  =  A  f£l  PV  <84) 

For  a  change 

72  -  /!  =  A  7r^-r(p2F2  -  PI^X)  (85) 


ENTROPY  OF  GASES 

For  reversible  lines 

,         dq 
ds  =  -~ 

hence  from  (48),  (49),  and  (50): 


s*  -  81  =    Ccv~+    CA  PTdv  =  cvloge^  +  AB\oge^  (86J 
s2  -  si  =  cp  loge  ^  —  AE  loge  -^  (87) 

S2   -    Si    =    Cp  loge  ~  +  C^  10S«    ~  (88) 

These  changes  in  entropy  depend  only  on  the  states  at  the 
beginning  and  the  end  of  the  path. 

CYCLES  AND  CROSS  PRODUCTS 

Cycles  are  the  paths  showing  the  changes  in  the  properties  of  a 
substance  as  it  undergoes  a  change.  They  are  often  shown  on 
the  pv  plane  as  in  Fig.  9.  A  simple  cycle  is  one  made  up  of  two 


38 


HEAT  ENGINEERING 


pairs  of  similar  lines.     A  cycle  of  four  or  more  different  lines  is 
a  complex  cycle. 

The  cycles  of  engines  using  perfect  gases  have  certain  properties 
provided  that  they  are  made  up  of  two  pairs  of  polytropics.  If 
the  simple  cycle  is  made  up  of  polytropics,  pvn  =  const., 

FiF3  =  V2V*  (89) 

Pips    =  Pzpi  (90) 

T,T3  =  T2Ti  (91) 


pv  n  -  Const 


-  Const. 


FIG.  9. — Simple  cycle  of  polytropics. 
This  is  shown  as  follows: 


(89) 


fCM\\ 
PIPS    =    P2P4:  CyW 

If  the  cross  products  for  pressures  and  for  volumes  are  equal, 
those  for  temperature  must  be  equal. 


T^  nn    — —  T^  '/^  f  Q~l ) 

This  latter  is  true  for  perfect  gases  only.     For  cycles  of  any 
substance  the  cross  products  of  pressures  or  volumes  are  equal 


FUNDAMENTAL  THERMODYNAMICS  39 

from  the  geometry  of  the  figure  but  the  cross  products  of  tem- 
peratures are  only  true  for  perfect  gases. 

SATURATED  STEAM  AND  OTHER  VAPORS 

A  vapor  is  a  gaseous  condition  of  a  substance  near  its  point  of 
liquefaction. 

When  a  vapor  is  in  contact  with  its  liquid  it  is  said  to  be 
saturated. 

The  characteristic  equation  for  steam  (due  to  Bertrand)  is 

T 

log  p  =  log  k  -  n  log  T  _  b  (92) 

p  =  pressure  in  pounds  per  sq.  in. 
T  =  absolute  temperature  in  deg.  F. 
n  =  50 

32°  F.  to  90°  F.          90°  F.  to  237°  F.       238°  F.  to  420°  F. 
b  =  140.1  b  =  141.43  b  =  140.8 

log    k  =  6.23167  log  k  =  6.30217         log  k  =  6.27756 

For  other  vapors  Goodenough  gives  the  Dupre"-Hertz  formula 
with  the  value  of  constants  from  Bertrand. 

C 

log  p  =  a  -  b  log  T  -  j,  (93) 

p  =  pressure  in  millimeters  of  mercury 
T  =  absolute  temperature  in  degree  C. 

a  b                           c 

Water  17.44324  3.8682  2795.0 

Ether  13.42311  1.9787  1729.97 

Alcohol  21.44687  4.2248  2734.8 

Sulphur  dioxide  16 . 99036  3 . 2198  1604 . 8 

Ammonia  13.37156  1.8726  1449.8 

Carbon  dioxide  6.41443  -0.4186  819.77 

These  equations  are  rarely  used  since  tables  of  the  properties 
of  vapors  have  been  constructed  giving  the  pressures  and  corre- 
sponding temperatures.  From  the  characteristic  equations  of 
saturated  vapors  it  is  seen  that  the  pressure  and  temperature 
are  independent  of  the  volume.  When  heat  is  added  to  1  Ib. 
of  liquid  at  32°  F.  and  of  volume  v'  it  is  found  that  the  volume  in- 
creases a  very  slight  amount  so  that  dv  may  be  considered  as 
zero,  giving  dq  =  cvdt,  or  better  cdt  to  include  the  slight  amount 
of  work. 


40  HEAT  ENGINEERING 

The  value  of  c  has  been  determined  experimentally  for  dif- 
ferent temperatures,  hence  if  these  be  plotted  as  functions  of  t, 
the  area  beneath  the  curve  will  give  the  value  of  the  integral 

q'  =    \    cdt  (94) 

This  is  called  the  heat  of  the  liquid.  It  is  the  amount  of  heat 
to  raise  1  Ib.  of  liquid  from  32  to  some  temperature  t.  For 
water  it  is  found  graphically  by  plotting  c  but  for  other  sub- 
stances the  quantity  is  found  by  empirical  equations  of  the  form 

q'  =  a  +  bt  +  ct2  +  .    .  |  (95) 

c  =  ^JL  =  5  _|_  2ct  +  Zdt2  +  .    .    .    .  (96) 

After  the  liquid  has  been  heated  to  the  temperature  corre- 
sponding to  the  pressure,  the  addition  of  heat  causes  the  liquid 
to  boil  and  some  of  the  liquid  is  changed  into  vapor.  When 
all  is  changed  into  vapor  the  volume  of  1  Ib.  is  v"  so  that 
the  volume  has  been  changed  by  the  amount 

v"  —  v'  =  change  of  volume. 

If  x  represents  the  amount  of  1  Ib.  which  has  been  changed 
into  steam  it  is  called  the  quality,  (1  —  x)  is  the  amount  which  re- 
mains liquid.  The  volume  of  the  1  Ib.  of  mixture  is 

v  =  v'  +  x(v"  -  v')  =  (1  -  x)vf  +  xv"  (97) 

V  =  M[(l  -  x)v'  +  xv"}  (98) 

Since  v'  is  small  and  since  (1  —  x)  is  small  in  most  cases, 
(1  —  x)v'  may  be  neglected,  giving 

V  =  Mxv"  (99) 

The  amount  of  heat  added  to  change  1  Ib.  of  liquid  from 
liquid  at  the  boiling  point  to  vapor  at  this  temperature  is  called 
the  heat  of  vaporization  and  is  represented  by  r. 


r  =  fdq  = 

Now  T  is  constant  and  (^7)  is  independent  of  v,  hence 


(100) 


FUNDAMENTAL  THERMODYNAMICS  41 

This  equation  may  be  used  to  find  r  if  v"  —  v'  is  known. 
Since  r  is  usually  found  by  an  empirical  equation,  equation  (100) 
is  used  to  compute  (v"  —  v'). 

For  steam 

r  =  970.4  -  0.6550  -  212)  -  0.000450  -  212)2    (101) 

The  total  heat  added  to  1  Ib.  of  liquid  at  32°  F.  to  make  it 
into  steam  at  a  given  temperature  is  called  the  total  heat,  q"  or 

q"  =  q'  +  r  (102) 

For  steam 

q"  =  1150.4  +  0.350  -  212)  +  0.000333(t  -  212)2  (103) 

For  other  substances  the  equations  may  be  found  in  tables  of 
properties. 

Having  r,  v"  —  v'  may  be  found;  since  by  (92), 

dp  _        nbp 
dt   :=  T(T  -  b) 

v"  -  „'  = -£ 

A  T_J*P_  (104) 

T(T  -  b) 

If  the  volume  is  changed  by  v"  —  v'  at  a  pressure  p,  the 
external  work  expressed  in  heat  units  is 

A  work  =  Ap(v"  -  v')  =  ¥  (105) 

The  internal  heat  of  vaporization  is  therefore 

P  =  r  -  V  (106) 

This  is  computed  and  placed  in  the  tables. 

If  the  steam  or  vapor  is  such  that  no  liquid  is  present,  x  =  1 
and  the  vapor  is  dry.  However,  if  there  is  some  liquid,  x  is  not 
unity  and  the  vapor  is  wet.  In  either  case  it  is  a  saturated  vapor 
as  the  pressure  and  temperature  are  related  by  means  of  the 
characteristic  equation  of  saturated  vapor.  When  x  is  not 
unity  the  total  heat  is 

q."  =  q'  +  xr 
INTRINSIC  ENERGY 

There  is  practically  no  work  when  the  liquid  is  heated,  hence 
the  quantity  of  heat  q'  remains  in  the  body  and  when  steam  is 
made  the  change  of  intrinsic  energy  is  given  by 

A(u  -  u32)  =  q'  +  r  -  Ap(v"  -  v'} 
=  <?'+/> 


42 


HEAT  ENGINEERING 


Since  the  intrinsic  energy  may  be  measured  from  32°  F.,  this 

atr   V»o   IXTTM •ff on 


may  be  written 

Au  =  q'  +  p,  or  AU  =  Jlf  fa'  +  p) 
For  wet  vapor 


\'  +  xp). 


=  q'  +  xp,  or 

HEAT  ON  A  PATH 
If  heat  is  added  to  a  vapor  on  some  path  as  in  Fig.  10 

q  =  A(U»  -  U,] 


(107) 
(107') 


=  AM  [q'2 


.J  pd 


FIG.  10.  —  Path  on  pV  plane. 

The  area  beneath  the  curve  is  the  value  of  the  integral.     This 
may  be  of  any  of  the  forms  given  earlier. 
To  find  x  the  formula 


<108) 


is  used. 


HEAT  CONTENT 


At  times  steam  tables  and  charts  give  the  value  of  the  heat 
content  and  the  method  of  using  this  must  be  clearly  understood. 

i  =A(u  +  pv") 

=  qr  +  r'  -  Ap(v"  -  v')  +  Apv" 
=  qr  +  r'  -  Apv\ 

Now  Apv'  is  a  very  small  quantity,  hence  i  and  q"  are  almost 
the  same.     If,  however,  i  is  given 

i  -  Apv  =  Au  (109) 

so  that  for  A(u*  —  HI) 


may  be  written  and  problems  solved. 


FUNDAMENTAL  THERMODYNAMICS  43 

ENTROPY 
The  entropy  change  when  the  liquid  is  heated  is 

CT     dt        CT 

St  -  S32  =          c  -  =    i      cd(loge  T)  =  s'  (110) 

1/491.6    -*  J|91.6 

This  entropy  of  the  liquid  is  found  graphically  as  the  area 
under  a  curve  of  c  plotted  against  log  T. 

When  the  liquid  at  the  boiling  point  is  vaporized 


The  total  change  of  entropy  from  liquid  at  32°  F.  into  dry 
vapor  is 

Since  32°  F.  is  the  datum  plane  of  reference 

s"  =  s'  +  £  (112) 

For  wet  vapor  s"  =  s'  +  ^  (1120 

DIFFERENTIAL  EFFECT  OF  HEAT 

When  a  mixture  of  liquid  and  vapor  has  a  differential  amount 
of  heat  added  to  it,  an  amount  dx  of  liquid  is  vaporized,  and  the 
liquid   (I  —  x)  and  the  vapor  x  are  raised  dt  degrees.     If  c'  is 
the  specific  heat  of  the  liquid  and  c"  that  of  the  vapor, 
dq  =  c'(l  -  x)  dt  +  c"  X  dt  +  rdx 


Now  ds  =  ^  =  ^-  ~-^P      -dt  +  ^dx 

This  is  an  exact  differential,  hence 

5  /c'(l  -x)+  c"x\  5 


x\          J_ 

}T~8T\T 


T       T  =    T  5T       T2 
c"  =  c'  -  £  +  §  (113) 

as                                         c'  =  dT 
and  :=  —  ~f~  ~T~~ 

„      dq"       r  /iio/\ 

•  •c    =  -Jtfr  -  m  (H3  J 


44  HEAT  ENGINEERING 

Since  q"  =  a  +  &(«  -  212)  -  c(t  -  212)2 

^  =  6  -  2c(<  -  212) 

may  be  substituted  and  c"  may  be  found. 

SUPERHEATED  VAPOR 

If  the  saturated  vapor  is  taken  away  from  its  liquid  so  that 
additional  heat  will  not  vaporize  any  liquid,  the  addition  of 
heat  will  cause  the  temperature  to  rise  above  its  saturation 
value  if  the  pressure  is  kept  constant.  This  vapor  is  known  as 
superheated  vapor. 

The  equation  for  superheated  steam  as  reduced  by  Good- 
enough  from  the  work  of  Knoblauch,  Linde  and  Klebe  is 


p  =  pressure  in  pounds  per  square  inch 
v  =  volume  of  1  Ib.  in  cu.  ft. 
B  =  0.5963 
log  m  =  13.67938 
n  =  5 
c  =  0.088 
a  =  0.0006. 

For  other  substances  the  equation  of  Zeuner  is  used 

pv  =  BT  -  cpn  (115) 

For  high  temperatures  Mallard  and  Le  Chatelier  and  Langen 
give  for  the  specific  heat  of  superheated  steam  at  constant 
pressure 

cp  =  0.439  +  0.000239*  (116) 

but  from  (56)  according  to  Goodenough 

(tep\  AT  — 

\5p)T-  2  dt* 

From  (114)  8v       B        mn  ..  . 

8T  =      +T^(l+ap) 

dc\      1 


Amn(n  - 


FUNDAMENTAL  THERMODYNAMICS  45 

Using  (116)  as  the  form  of  expression  for  <f>T 

Amn(n  — 


c 


m          mnn  —          /         a 
=  a  +  0T  +  -      yUi     ~P(l  +  2P 


From  the  results  of  Knoblauch  and  Mollier,  Goodenough  re- 
duces for  a  and  0  the  values 

a  =  0.367 
j8  =  0.0001 

giving         cp  =  0.367  +  0.000177  +  p  (I  +  0.0003p)  ~         (117') 

log  C  =  14.42408 

7?  =  pounds  per  square  inch 
T  =  degrees  absolute  F. 

This  value  of  cp  agrees  with  the  results  of  experiment  by 
Knoblauch  and  Mollier.  Values  of  this  are  shown  in  Fig.  11  as 
given  by  Goodenough. 

The  value  of  the  specific  heat  of  other  superheated  vapors 
is  given  below  and  these  are  usually  taken  as  constants  although 
this  is  not  true. 

SPECIFIC  HEATS  AT  CONSTANT  PRESSURE  AND  K'S 

Superheated  ammonia  vapor  0.536  k  =  1.32 

Superheated  sulphur  dioxide  0.1544  k  =  1.26 

Superheated  carbon  dioxide  0.215  k  =  1.30 

Superheated  chloroform  0.144  k  =  1.10 

Superheated  ether  0.462  k  =  1.03 

If  now  1  Ib.  of  liquid  at  32°  F.  is  heated  and  finally  changed 
into  superheated  vapor  at  temperature  Tsup.  the  amount  of  heat 
required  is  given  by 

f*TtUp. 

g,'"  =  q'  +  r  +    I  cpdt  (118) 

J  T.at. 

The  last  term  is  found  graphically  or  analytically  for  different 
pressures  and  different  amounts  of  superheat  and  tabulated  or 
plotted  in  charts  so  that  q"'  may  be  known. 


The  value  of   I  cpdv  may  be  written  as 

. 

(mean  cp)  (T8up.  -  Tsat.) 


HEAT  ENGINEERING 


0.450 


0.425 


200  300  400 

Degrees  of  Superheat 
FIG.  11. — Specific  heat  of  superheated  steam. 


500 


FUNDAMENTAL  THERMODYNAMICS  47 

The  value  of  mean  cp  is  given  by 

rT.UP. 
I  Cpdt 

mean  cp  =  T  jTtai'  T 

•*-  sup.  -*-  sat. 

The  values  of  mean  cp  have  been  compared  and  plotted  in  the 
form  of  curves  by  Goodenough.  From  these  curves  the  table 
on  p.  48  has  been  constructed: 

The  volume  of  this  superheated  steam  is  computed  by  (114) 
and  tabulated  or  plotted. 

The  heat  remaining  is 

Au  =  qf"  -  Ap(v"f  -  v') 
But     Au  =  i  -  Apv'".     (109) 
/.  i  =  q'"  +  Apv' 
i  =  q"'  practically. 

The  entropy  of  the  superheated  vapor  measured  from  liquid 
at  32°  F.  is 


(119) 

Tsat. 

This  last  term  is  found  analytically  or  graphically,  but  in  most 
cases  these  have  been  tabulated  for  steam  so  that  values  of 
s'"  may  be  found  for  definite  conditions. 

The  various  quantities  for  saturated  steam  mixtures  and  super- 
heated steam  may  be  found  and  these  properties  are  often  made 
into  charts.  The  quantities  p,  v,  t,  s,  u,  i,  x  are  all  dependent  on 
the  state  and,  if  known,  fix  the  state.  Usually  any  two  of  them 
fix  the  state,  so  that  these  two  could  be  used  for  the  coordinates 
of  a  diagram.  As  has  been  shown  before,  T  and  S  could  be  used, 
and  in  this  the  area  under  a  line  represents  the  heat  added  from 
the  outside  if  the  line  is  reversible  or  the  heat  added  from  the 
outside  together  with  friction  if  the  line  is  non-reversible. 

Other  coordinates  such  as  i  and  s  have  been  proposed  by 
Mollier.  This  is  known  as  a  Mollier  diagram.  These  will  now 
be  explained. 

T-S  AND  I-S  CHARTS 

In  Fig.  12  the  coordinates  T-S  are  used  and  since  heat  is  added 
to  the  water  at  32°  F.  the  line  of 


,  r  dt 

>  =  ]     CT 

J  491.6 


48 


HEAT  ENGINEERING 


-jadns  jo 


Ss 


000 


co  co  co 
odd 


odd 


1C  lO  «C  1C  UJ 
OOOOO 


CO  CO  1C  lO 


MOOOcO 

IC»C  TjH  T}< 

1C  «O  10  1C 

dddd 


O  O  C5  GO  00 
OOOOO 


1C  1C  iC 
OOO 


C4}  00  ^t4    I   CM  00  CO 

t^  CO  CO    I   CO  1C  iC 

odd  I  odd 


o 


I  C5COCO 

I   iC  1C  iC 

odd 


OOiOfN 

odd 


iOiO»O»OiO 

doodd 


COOOOiOCO 

d  d  d  d  d 


S888S? 

ddddd 


oco  co  •**<  co 

COIN  <N  <N  CM 
1C  1C  iO  1C  1C 

ooooo 


1C  1C  "5 

odd 


'*Tt!'*. 
odd 


CO  CO  CO 

••*••*  TfH 

OOO 


ooo  i  ooooo 


ooo  ooooo 


'^!rt!T*! 

odd 


1333 
I  odd 


o 


'*•*'* 

odd 


888 


HI 

ooo 


co  co  co 
odd 


O  O  O  GO  GO 

t-,t>.  CD  CD  O 


ooooo 


ooooo 


-*  Tt<  CO  CO  CO 

ddddd 


co  co  co  co  co 

rf<  rf  Tf  T}<  ^ 
OOOOO 


00  CO  00  T^  i-H 
t>  t>.  CO  CO  CO 
1C  iC  1C  1C  1C 

ddddd 


lOOCOtNO 

CO  CO  iC  1C  Tf 

ddddd 


ooooo 


00»CtNO5l^ 
COCO  CO  CN  CM 
1C  «3  1C  "3  fJ 


1C  U3  1C  1C  >C 

ddddd 


ooooo 


OOCO»CTt<CO 

CO  GO  00  00  GO 


ooooo 


OOOO  t~  CO 

ddddd 


CO  CO  CD  CD  CD 

ddddd 


»CIC 

Tj<  -tf 

dd 


1C  iC  O  1C  iC 

ooooo 


00105OO 

d  d  d  d  d 


ooooo 


OOO^Cjj 

ddddd 


-^  OcO 

ic  10  •* 

1C  10  iC 


iC  iO  iC  iO 

oooo 


OOOO 


1C  CO  OOO 
10  3  tfj  1C 

dddd 


ssss 

oooo 


(NOOOO 

O  O  GO  GO 

dddd 


00000000 

T)<-*rt(Tj( 

dddd 


ic  ic  ic  co 

CD  CD  CD  CD 


CO^HO 

co  co  co 

1C  >C  »C  i 


5S   S3 

do  I  do 


CM  <N    i   (N  (N 

1C  1C       1C  1C 


(NOOiQO 
(N  <N  1-1  i-H 
1C  »O  1C  1C 


00 


00 


do  I  oo 


t^cOiOiC  "*•* 

OOOO  I  O  O 

1C  1C  lOiO  iC  1C 

ddo'd  do 


oooo i oo 


!-H<N  (N  (N 

00  00  GO  GO 


SI-HN  (M 
iCiCiC 


10  iC  COCO  t- 


OOOOO   I  OOOCO 


5533 

dddd 


CO  t^  OOO 

Tf  Tf<  rj<  T* 

"*.  "*  "*.  "^ 

dddd 


00 


S3 
do 


00  O       CM  1C 

do  !  do 


iO  QiO 
C-l  1C  h* 

CO  CO  CO 


FUNDAMENTAL  THERMODYNAMICS 


49 


starts  at  32°  F.  or  491.6°  abs.  F.  This  line  is  shown  as  the  liquid 
line  in  Fig.  12.  The  various  values  of  sr  for  different  values  of  T 
from  the  steam  tables  are  plotted.  From  points  on  the  liquid 

line  the  distances  ^  for  those  temperatures  are  laid  off  giving  the 
saturation  line.  It  is  evident  that  these  two  lines  approach 


432  F 


FIG.  12.  —  TS  diagram  for  TS  analyses.        FIG.  13.  —  TS  diagram  areas. 

and  meet  at  the  critical  temperature  (at  which  the  superheated, 
saturated  and  liquid  states  coincide),  since  at  this  point  r  =  0. 

Since  dq  =  Tds 

and  dq  =  cdt 


or  the  specific  heat  is  the  subtangent  from  any  point  on  a  curve 
on  the  T-S  plane.  In  Fig.  12  it  will  be  seen  that  while  the  specific 
heat  of  the  liquid  is  positive,  the  specific  heat  of  saturated  steam 
is  negative. 

The  areas  beneath  lines  on  which  there  is  no  friction  are 
equal  to  the  heat  added.     As  shown  in  Fig.  13,  the  area  beneath 


50 


HEAT  ENGINEERING 


the  liquid  line  is  q'  while  that  beneath  the  line  from  the  liquid  line 
to  the  saturation  line  is  r.  It  must  be  remembered  distinctly 
that  steam  tables  and  diagrams  assume  that  steam  is  made 
by  keeping  the  pressure  constant,  heating  the  water  from  32°  F. 
to  the  boiling  point,  boiling  to  dry  vapor  and  then  superheating  at 
constant  pressure.  If  superheating  takes  place  the  temperature 
will  rise  by  the  degrees  of  superheat  and  the  entropy  will  increase 
by 


(cjdt 
)    T 


the  line  will  therefore  take  the  shape  cd.     The  broken  line  abed 
is  a  line  of  constant  pressure.     The  area  beneath  cd  is 


If  the  steam  is  wet  the  change  of  entropy  from  the  liquid 

XT 

condition  is  ™-  ,  and  hence  if  point  e  is  so  selected  that 

be 


the  point  e  represents  the  condition  of  quality  x  since  be  =  ™. 

If  now  for  various  temperatures  (and  consequently  saturation 

pressures)  various  points  be 
found,  having  the  same 
value  of  x,  a  line  of  constant 
quality  or  constant  steam 
weight  may  be  obtained  as 
shown  in  Fig.  14. 

In  the  superheated  region 
this  is  replaced  by  lines  of 
°\  constant  amount  of  super- 
heat.  In  this  case  points 
on  various  pressure  lines  at 
the  same  number  of  de- 
grees above  saturation  are 
connected. 


FIG.  14. — Lines  on  TS  plane. 


If  the  value  of  i  be  computed  for  point  e  and  this  is  made  equal 
to  the  i  at  some  other  temperature  the  quality  at  this  point  may 
be  found  and  from  it  the  position  of  e  on  that  line. 

q'  +  Xr  =  g'i  +  X1rl 

q'  +  xr  -  q\ 


FUNDAMENTAL  THERMODYNAMICS 


51 


If  a  number  of  points  are  found,  a  line  of  constant  heat  content 
is  obtained.     In  the  superheated  region 


q'  +  r  +  I  cpdt  =  q\  +  n  + 


and  the  values  of  the  degrees  of  superheat  at  the  second  pres- 
sures are  found  to  give  the  same  heat  content  as  at  the  first  point. 
Such  points  form  the  line  of  constant  heat  content. 
It  is  known  that 

u  =  q'  +  xp  or  u  =  i  —  Apv 

Hence  if  the  value  of  u  at  point  e  is  equated  to  the  expression  at 
another  pressure  or  temperature,  the  quality  x  or  the  degrees  of 
superheat  may  be  found  for  the  second  pressure  or  temperature 
and  this  fixes  the  position  of  the  point.  Connecting  a  series 
of  these  the  line  of  constant  intrinsic  energy  is  found. 
If  the  volume  of  the  point  e  is  found  as 

v  =  xv" 

and  equated  to  x\vr\  for  a  different  pressure  the  value  of  x  may 
be  found,  and  from  a  series  of  these  points  a  line  of  constant 


1400 


1.20 


2.00 


FIG.  15. — Mollier  chart. 


volume.  In  the  superheated  region  the  long  formula  for  super- 
heated steam  would  have  to  be  used  and  by  equating  this  for  two 
pressures  the  degrees  of  superheat  at  the  second  point  could 
be  found  and  in  this  way  the  curve  could  be  determined. 

The  lines  of  constant  S  and  constant  temperature  are  vertical 
and  horizontal  lines. 


52  HEAT  ENGINEERING 

A  diagram  such  as  Fig.  14  is  used  in  practice  in  which  lines  of 
constant  i,  x}  v,  deg.-sup.,  s  and  t  are  drawn  at  equal  distances 
apart. 

In  the  Mollier  chart,  Fig.  15,  similar  methods  are  used  to  com- 
pute various  points.  The  lines  shown  are  constant  quality  and 
pressure  on  i-s  coordinates  and  to  this  diagram  should  be  added 
lines  of  constant  volume. 

For  a  definite  pressure 

.    .    xr          ,        r        C    dt 
s  =  s '  +  -^   or  s'  +  ^  +  I  cp  -y 

and  i  =  q'  +  xr—  Apv'  or   q'  +  r  -f    I  cpdt  —  Apv' 

and  for  different  values  of  x  or  degrees  of  superheat  at  the  same 
pressure  the  values  of  i  and  s  may  be  found  giving  a  line  of  con- 
stant pressure. 

If  this  is  done  for  several  pressures  and  then  the  points  of  the 
same  quality  are  connected  the  lines  of  equal  quality  and  equal 
pressure  are  found.  If 


or  the  equation  for  superheated  steam  is  used  the  conditions  at 
different  pressures  for  the  same  volume  may  be  found  and,  from 
these,  lines  of  constant  volumes.  Fig.  15  shows  this  chart. 

Having  these  charts  and  formulae,  the  various  lines  may  be 
discussed.  On  all  lines  the  important  things  are  the  work,  heat 
and  change  of  entropy. 

HEAT  ON  PATHS 
If  Fig.  16  represents  any  line  the  heat  added  on  it  is  given  by 

JQ  =  u*  -  Ui  +  \pdv 

r  J 

I  pdv  =  work 

To  find  this  analytically  the  equation  of  the  curve  must  be 
known  and  if  this  is  not  known  the  area  beneath  the  curve  must 
be  found  by  a  planimeter  and  the  graphical  method  used  to 
evaluate  the  integral.  If  the  curve  is  concave  upward,  as 


FUNDAMENTAL  THERMODYNAMICS 


53 


shown  dotted  in  Fig.  16,  it.  may  be  assumed  to  be  of  the  form 
pvn  —  const.,  in  which  case 


pdv  = 


1  -  n 


for  all  values  of  n  except  unity;  n  must  be  known  for  this  and 
from  the  pressures  and  volumes  at  1  and  2  it  is  found  as  follows  : 


Pi        Z§\ 
p*  "  \VJ 


n 


(121) 


log 


Fi 


Of  course  this  is  the  value  of  n  of  a  curve  of  this  form  passing 
between  the  end  points  1  and  2, 
but  it  may  not  pass  through  the 
other  points. 

The  specific  volumes  at  points  1 
and  2  are  now  found  by  dividing  the 
volume  by  M  after  which  these  spe- 
cific volumes  and  pressures  are  used 
on  charts  or  in  tables. 

U  =  M(i  -  Apv) 
or  U  =  M(q'  +  xp) 

To  find  this,  i  or  x  must  be  known. 
If  from  tables  or  charts  i  may  be 
found,  then  the  first  formula  is  the  one  to  use  as  it  applies  to  sat- 
urated and  superheated  conditions.  If,  however,  the  table  does 
not  go  low  enough  x  is  found  by 

Fi 

Xl  ~  Mv\ 
V, 


FIG.  16. — Path  on  pV  plane. 


Of  course  M  must  be  given  in  the  problem;  v"  is  taken  from 
the  steam  tables. 

Having  U2  and  Ui  their  difference  is  known  and  from  this  and 
work  the  heat  is  found  by  addition. 

The  change  of  entropy  is  determined  by  finding  the  entropies 


54  HEAT  ENGINEERING 

at  1  and  2  from  charts  or  tables  or  if  these  cannot  be  used  they 
must  be  computed 

/    ,    xr 

8   =  S     +    y- 

The  entropy  change  is  then 

S%  —  Si  =  M(s2  —  Si) 

This  is  for  any  line.  For  the  adiabatic,  the  heat  is  zero,  the 
entropy  change  is  zero,  and  the  work  is  equal  to  the  change  in 
intrinsic  energy.  The  entropy  is  constant  on  this  line  and  this 
gives  the  equation  of  the  adiabatic 

a'     4-  r  -   o'     4_  /*•  fl  99^ 

6  1  -f-  J>1  rp     '      *  2  ~T  *C2    rp  \*-"4) 

r,  dt 


Or  =  S'2  +  TfT  + 

1  2 


L  sup- 

*, 

r..,,. 


Knowing  the  first  point  through  which  the  curve  must  be 
drawn,  the  unit  volume  is  found  by 

V 

v  =   ~Tur  ' 

M 

From  tables  or  charts  the  quality  is  found  for  this  specific 
volume.  If  this  cannot  be  found  on  account  of  the  limits  of 
these  tables 

v 

**> 

The  entropy  of  the  first  point  being  known,  the  quality  at  the 
second  point  is  found  by  (122)  and  from  this  the  volume,  if  needed, 
is 

V  =  Mxv"  or  Mv  ' 

If  v  can  be  found  from  tables  or  charts,  it  is  easier  to  get  it 
from  the  same  entropy  column  or  line.  The  work  is  then  given 
by 

Work  =  Ui  -  Ut  =  M(ui  -  u2) 
u  =  i  —  Apv  or  q'  +  xp 

At  times  the  difference  between  the  u's  would  lead  to  such  an 
error  that  it  is  better  to  find  Vz  for  a  given  p2  and  then  compute 
n  by  the  log  formula  (121)  and  find  work  by 

Work  = 


FUNDAMENTAL  THERMODYNAMICS  55 

The  isodynamic  is  a  line  of  constant  U. 
Ul  =  U2 


Now  Ui  =  i  —  Apv  or  q'  -f  xp 

.'.  ii  —  ApiVi  =  iz  —  ApzVz  or  q\  +  Xipi  =  q'2  + 

(123) 

Given  the  pressure,  volume  and  weight  at  the  first  point,  the 
quality  is  found  after  determining  the  specific  volume  by  using 
the  tables  and  charts  if  possible;  or  by  using  the  formula  for 
superheated  steam  or 


The  value  of  Ui  may  then  be  found  and  this  is  equated  to  uz 
for  the  second  pressure  from  which  the  second  quality  may  be 
computed  and  then  the  volume  is  found  by 

Vz  =  Mxzv"z 

The  heat  in  this  case  is  equal  to  the  work  since  there  is  no 
change  of  intrinsic  energy.  The  work  is  found  by  finding  n  by 
formula  (121)  and  then 

Work  ,  PiZ^PiZ. 

1  —  n 
The  values  of  s2  and  Si  are  found  and  then 

St  -  Si  =  M(SZ  -  si) 

The  isothermal  in  the  saturated  region  is  the  same  as  the  con- 
stant pressure  line  and  on  this 

Q  =  M(xz  -  Xl)r  (124) 

Work  =  M(xz  -  xi)t  (125) 

S2  -  Si  =  M(xz  -  x,)  £  (126) 

In  the  superheated  region  the  curve  approaches  the  rectangular 
hyperbola  and  it  would  be  difficult  to  solve  problems  on  this  line 
if  tables  were  not  available.  In  this  case  the  formula  for  super- 
heated steam  would  have  to  be  used  to  get  the  volumes  or  if 
tables  or  charts  were  used  the  volumes  could  be  found  by  finding 
the  degrees  of  superheat  at  various  pressures  by 

Degree  superheat  =  T  —  Taat, 


56 


HEAT  ENGINEERING 


and  this  would  fix  i,  v  and  s.     Then 


Work 


Pi7 


1  -  n 

Q  =  U2  —  Ui  +  work 
S2  —  Si  =  M(SZ  —  Si) 
On  the  constant  volume  line 

Work  =  0 

7 


\i ; 


Uz  -  Ui  =  Q  =  M[q'z  +  Z2p2  -  (q'i  +  XiPi)] 
Sz  —  Si  =  M(SZ  —  Si) 

FLOW  OF  FLUIDS 

If  a  fluid  flows  through  an  orifice  in  which  there  is  a  change  in 

,  section  there  is  a  drop  in  pressure  and 

as  a  result  there  is  a  change  in  velocity. 
In  Fig.  17  consider  the  sections  1  and 
2.  If  the  areas  are  represented  by  FI 
and  FZ,  the  pressures  by  pi  and  pz}  the 
specific  volumes  by  Vi  and  t;2,  the  veloc- 
ities by  Wi  and  wz  the  following  consid- 


FIG.  17.— Orifice  for  flow    erations   must  hold  when  M  pounds  of 


of  fluid. 


substance  flow  per  second. 


Kinetic  energy  at  1  in  ft-lbs. 

Internal  energy  at  1  =  Mui 

Work  done  in  pushing  substance  along 

Hence  total  energy  at  1 
Total  energy  at  2 


=  M 


Now  if  there  is  any  heat  added  between  these  two  points,  say 
MJq,  the  energy  at  2  must  equal  that  at  1  plus  MJq.  Of  course 
if  Jq  is  taken  away  the  sign  would  change. 


Hence       ^— 


+ 


+  Jq  =  ^  +  uz  + 


or 


since 


T"      ~    =    ^  +  ^1  + 

Q 

=  Jq  +  J(ti 


~    (^2   + 


(127) 


(128) 


If  now 

becomes 


FUNDAMENTAL  THERMODYNAMICS       Q\        57 
is  so  large  that  Wi  is  small  and  if  q  =  0,  the  formula 

2 
7TT    =   «f(*l  ~~  *2) 


or 


(129) 
(130) 

Since  q  =  0  this  action  is  adiabatic  but  in  the  case  that  there 
be  internal  friction,  1  and  2  are  not  on  points  on  an  isoentropic 

line  but  a  line  passing  to  the  right  of 
this  line  on  the  T-S  plane.  In  other 
words,  iz  is  greater  than  it  would  have 
been  if  there  had  been  no  friction. 
The  amount  of  this  increase  in  iz  is 
usually  found  by  assuming  that  2  has 
the  same  entropy  as  1  and  then  in- 

U  P 


FIG.  18.— TS  diagram  for  flow 
of  fluids. 


FIG.  19. — pV  diagram   for   flow   of 
fluids. 


stead  of  using  *i  —  22  for  these  points,  only  a  fraction  of  this  is 
used.     In  Fig.  18  abcde  is  equal  to  ii,  since 

V 


abcf  =  q' 
cdef  =  xr 
abcde  =  q'  + 
abg2e  =  i% 
i  —  iz  =  cd2g 


xr  =  i  —  Apv' 


This  assumes  no  friction,  hence  d  and  2  have  the  same  entropy. 
By  experiment  the  heat  used  in  friction  is  found  to  be  y(ii  —  i2), 
hence  this  must  be  subtracted  to  get  the  amount  of  heat  left  to 
give  the  gain  in  kinetic  energy.  Hence 

(131) 


-  y) 

ii  and  i2  are  for  points  on  the  same  entropy  line.     In  Fig.  18  the 
area  cd2g  is  cut  down  by  the  amount  hd2k  which  .isequal  to 


58  HEAT  ENGINEERING 

y(ii  —  it).     This  heat  stays  in  the  substance  and  hence  at  exit  i 
is  not  the  iof.2  but  that  of  2'  which  is  fixed  by  making 


If  this  value  of  iv  could  have  been  determined  the  original 
formula  (130)  would  have  been  used.  It  is  because  iy  cannot 
be  found  that  this  method  of  using  a  portion  of  the  amount  for 
isoentropic  expansion  is  employed. 

It  must  be  remembered  that  the  shaded  area  less  the  friction 
loss  equals  the  gain  of  kinetic  energy. 

2  —      2 

-  y) 


The  quantity  z\  —  iz  is  the  same  as  the  area  behind  the  adia- 
batic  on  the  pv  plane  as  shown  in  Fig.  19. 

Area  a!2c  =  aleb  +  el2d  -  c2db 

=  J(ii  —  iz) 
On  account  of  friction  this  is  reduced  as  pointed  out  above. 

THROTTLING  ACTION 

Suppose  now  that  there  is  so  much  friction  that  there  is  no  gain 
in  kinetic  energy  although  there  is  a  drop  in  pressure;  this  is 
called  throttling  action.  In  this  case 

"7% ~"~     f-k ~~    \J    *~"~    «/  v^l    ~"~    vfyj 

20        2g 
or  t!  =  iz  (132) 

This  means  that  in  throttling  action  the  heat  content  is  con- 
stant. The  point  2'  of  Fig.  18  then  is  found  to  be  on  a  curve  of 
constant  heat  content.  The  horizontal  lines  of  the  Mollier 
chart  or  the  constant  i  curves  of  the  T-S  diagrams  are  throttling 
curves  on  these  diagrams. 

Since  in  perfect  gases 

Ji  =^rr  PiVi  =  j-^j  MBTl  (133) 

the  throttling  curves  for  such  are  curves  of  constant  temperature. 


FUNDAMENTAL  THERMODYNAMICS  59 

VELOCITY  OF  VARIOUS  SUBSTANCES 

For  short  tubes  and  orifices  the  friction  is  negligible  and  the 
action  may  be  considered  isoentropic  as  well  as  adiabatic. 


For  liquids,  since  the  temperature  change  is  slight, 
i  =  Apv  +  const. 


-  pz]V 
Since  Fi  =  V2 

F  =  l 

m 

[Pi  ~  P*]^  =  h 
w*  =   V20/i  (134) 

This  is  the  usual  formula  from  hydraulics  for  the  velocity  of  a 
liquid. 

k 
For  gases  JYi  = 


A; 


Since 

This  is  the  formula  for  velocity  and  is  a  maximum  when  pz  =  0. 

DISCHARGE  FROM  ORIFICES 

Now  Mv  =  volume  per  second  =  Fw 


This  is  zero  when  p2  =  PI  and  when  p%  =  0.     Of  course  the 
latter  is  unthinkable  because  there  will  always  be  some  weight 


60  HEAT  ENGINEERING 

when  pz  is  less  than  pi.  The  explanation  is  that  when  the  pres- 
sure p2  falls  the  pressure  which  exists  in  the  plane  of  the  orifice 
where  the  area  is  F  can  never  become  less  than  that  to  give  maxi- 
mum discharge.  To  find  the  condition  for  a  maximum  value 
of  M,  the  variable  part  of  the  expression  is  differentiated.  For 
a  fixed  pi  and  Vi 

,p,x_|       /p**±i 

w     w 

is  differentiated  with  regard  to  the  variable  p2  and  by  equating 
this  to  zero,  the  condition  for  a  maximum  is 

1  (137) 


When  k  =  1.4  for  gases  this  becomes 

7)2  =  0.5283pi  (138) 

while  for  steam,  k  —  1.135,  and 

7)2  =  0.574pi  (139) 

When  p2  becomes  less  than  the  critical  value  given  above  for  a 
short  mouthpiece,  the  pressure  at  this  point  of  area  F  remains 
at  the  critical  value.  Hence  the  discharge  is  constant  for  all 
values  of  p2  below  the  critical  value.  For  air  this  has  been  proven 
to  be  true  experimentally  by  Fliegner  who  proposes 

M  =  0.53^-^  (140) 

VT, 

when  7)2  <  0.53pi 


and  ^  .  Lo^  .  fet<a-zM.  (14i) 


when  pa  >  0.53pi 

Equation  (140)  is  really  a  reduction  of  (136)  by  substituting 
(138)  for  p2  and  reducing. 

For  steam  similar  experimental  results  have  been  found  by 
Napier.  Rankine  reduced  these  results  to  the  form 

M=^  "V.         (142) 

when  7>2  <  0.574pi  or  O.Gpi 


FUNDAMENTAL  THERMODYNAMICS  61 


when  p2  >  0.6/h 

The  general  equation  (131)  may  be  used  for  the  velocity  as 
shown  above  and  then  M  found  by 


after  v"%  is  found  for  the  condition  at  outflow. 


CHAPTER  II 
HEAT  ENGINES  AND  EFFICIENCIES 

A  heat  engine  is  any  machine  in  which  heat  is  used  to  furnish 
the  energy  for  the  production  of  mechanical  work.  As  examples, 
the  steam  engine,  the  steam  turbine,  the  gas  engine  and  similar 
machines  may  be  mentioned. 

To  make  heat  available  in  one  of  these  engines  it  must  be 
applied  to  some  substance  which  undergoes  changes.  These 
changes  form  a  cycle  and  the  changes  which  affect  the  properties 
of  the  substance  may  be  studied  on  planes  of  projection  by  show- 
ing the  successive  values  that  certain  properties  take.  The 
paths  of  the  change  which  the  substance  undergoes  are  known  as 
a  cycle. 

As  pointed  out  in  the  chapter  on  Fundamental  Thermo- 
dynamics Qi  heat  units  are  supplied  and  Q2  heat  units  are  re- 
jected, giving  Qi  —  Q2  units  of  work. 

All  of  these  machines  work  between  some  range  of  temperature, 
TI  to  T2,  and  consequently  the  availability  of  the  heat  is 

T  '        <» 

This  is  the  only  part  of  the  heat  which  could  be  turned  into 
work  and  therefore  represents  the  highest  possible  efficiency. 
It  represents  the  efficiency  of  the  Carnot  cycle  for  this  tempera- 
ture range  and  therefore  it  is  sometimes  spoken  of  as  the  Carnot 
efficiency  of  a  given  cycle.  Of  course  any  cycle  would  have  to 
be  a  Carnot  cycle  to  have  this  efficiency,  but  all  that  the  term 
means  with  reference  to  a  particular  cycle  is  the  maximum  value 
that  might  be  possible  for  this  cycle  working  between  TI  and  TV 
Calling  this  171 

Ti-T*  _          T*  (  . 

171  =    ~TT~      l  ~  T, 

If  Qi  is  the  theoretical  amount  of  heat  added  and  Q>2  the  theo- 
retical amount  rejected,  which  cannot  be  used, 

^    =   ,3  (3) 

Vl 

62 


HEAT  ENGINES  AND  EFFICIENCIES  63 

This  is  the  theoretical  efficiency  of  the  cycle. 

The  ratio  of  773  and  771  shows  how  close  the  efficiency  of  the 
theoretical  cycle  approaches  the  maximum  possible  efficiency  of 
the  cycle.  This  is  called  the  type  efficiency,  772 


If  this  is  nearly  unity  it  indicates  that  the  theoretical  cycle 
is  almost  equal  to  that  of  Carnot  and  hence  in  theory  the  cycle  is 
good  while  a  low  value  of  772  shows  that  the  cycle  is  a  poor  one 
theoretically.  For  example,  the  steam  engine  cycle  has  a  value  of 
772  above  0.90  while  in  the  case  of  the  gas  engine  772  may  be  less 
than  0.50.  These  indicate  that  an  improvement  may  be  ex- 
pected in  the  type  of  cycle  used  in  a  gas  engine,  although  for  the 
steam  engine  there  is  little  hope  of  bettering  the  cycle. 

If  for  an  amount  of  indicated  work  AWa  the  amount  of  heat 
Qa  is  actually  required,  the  actual  thermal  efficiency  775  is  given 
by 

AWa 
~Q7  =rib 

If  the  heat  supplied  per  unit  of  substance  used  is  qa,  be  it  coal, 
steam,  gas  or  air,  and  if  M  is  the  amount  of  substance  per 
indicated  horse-power  hour,  this  expression  becomes 

X  33000  X  60 


AWa       777 M  ~  www  ^  w      2546 
Qa  Mqa  ~  Mqa  ~  *5 

2546  B.t.u.  =  1  h.p.-hr. 
42.43  B.t.u.  =  1  h.p.-min. 

The  ratio  of  775  to  773  is  called  the  practical  efficiency,  774,  and 
shows  how  near  the  actual  efficiency  approaches  the  efficiency 
demanded  by  theory  and  a  low  value  of  this  means  that  there 
have  been  errors  in  actually  applying  the  cycle.  To  make  this 
term  larger,  jackets,  superheated  steam,  and  reheaters  have  been 
applied  to  steam  engines. 

175  ft~ 

* =  *  (6) 

If  the  mechanical  efficiency  776  is  the  ratio  of  the  output  to 
that  developed  within  the  machine  or  shown  by  the  indicator 
card,  this  efficiency  is  found  by 


64 


HEAT  ENGINEERING 
output 


indicated    work 


AW 
AW 


The  overall  efficiency  is  then  the  product  of  certain  of  these 
various  efficiencies. 

output 


heat    supplied 


(8) 


The  overall  efficiency  being  the  product  of  these,  the  efficiency 
may  be  increased  by  increasing  any  of  them.  In  cases  such  as  the 
gas  engine  771  is  so  great  that  although  772  and  776  are  small  the 
product  is  greater  than  that  of  the  steam  engine.  These  various 
efficiencies  will  be  investigated  for  different  machines. 


FIG.  20.— pV  and  TS  diagrams  of  cycles. 

The  quantity  AWa  may  be  found  from  the  area  of  the  cycle  on 
the  pv  plane  or  from  the  area  on  the  T-S  plane  in  Fig.  20,  if  the 
lines  are  reversible  lines.  Of  course  on  the  pv  plane  123456 
represents  positive  work  while  6571  represents  negative  work 
and  hence  723457  represents  the  network. 

On  the  T-S  plane,  123456  represents  Qi  if  there  is  no  friction, 
and  6571  represents  Q2  under  these  conditions,  hence  the  area 
of  the  cycle  represents  Qi  —  Q2  or  the  work  done.  If,  however, 
there  is  friction 

12345  =  Qi  +  H! 
and  6571  =  (-  Q2  +  #2) 

since  the  heat  removed  to  the  outside  from  5  to  7  is  more 
than  that  shown  by  the  area  on  account  of  the  heat  developed  by 
friction. 

23457  =  Q1  -  Q2  +  (Hi  +  #2)  =  AW  +  #1  +  H2 


HEAT  ENGINES  AND  EFFICIENCIES 


65 


or  the  area  is  greater  than  the  work  of  the  cycle.  In  any  case 
an  irreversible  line  makes  the  work  less  than  the  area  of  the  cycle 
on  the  T-S  plane. 

Looking  at  the  Carnot  cycle  of  Fig.  21,  or  the  cycle  of  Fig.  20, 
it  is  seen  that  the  efficiency  is 

2345        Ql  -  Q2 


123456  & 

If  in  the  Carnot  cycle  heat  is  not  added  on  34  but  on  some  other 
line  3'4  or  34',  the  heat  added  and  work  developed  are  decreased 
by  the  area  of  the  small  triangle. 

2345  -  33'4 


Hence 


Eff.  = 


123456  -  33'4 


Subtracting  the  same  thing  from  numerator  and  denominator 
of  a  fraction  less  than  unity  decreases  the  value  of  the  fraction. 
This  may  be  seen  by  remembering  that  the  effect  of  the  removal 
of  the  triangular  area  is  greater  on  the  smaller  area.  Hence 
the  efficiency  is  decreased. 


3/ 

8 

-->*^-^. 

4 

5" 

2 

n  —  •  —  ~  j 

5 

6                   S 

FIG.  21.  —  Conditions  for  maximum  efficiency. 

If  the  heat  were  removed  on  a  line  5  "2  of  Fig.  21  this  would  cut 
down  the  work  but  would  not  effect  the  heat,  hence  the  efficiency 
would  be  diminished  in  this  case. 

From  the  above  it  may  be  said  that  for  a  given  range  of  tem- 
perature heat  must  be  added  or  taken  away  at  constant  tempera- 
ture if  the  maximum  efficiency  is  to  be  obtained  and  since  this 
efficiency  is 


the  values  of  T2  and  T\  (the  limiting  temperatures  of  the  range) 
are  to  be  separated  as  much  as  possible. 

An  important  point  must  be  borne  in  mind.     Although  the 


66 


HEAT  ENGINEERING 


statement  above  is  absolutely  true,  it  does  not  follow  that  a 
cycle  in  which  heat  is  added  with  a  varying  temperature  finally 
reaching  a  high  value  may  not  be  more  efficient  than  one  in  which 
heat  is  added  at  a  constant  temperature  of  lower  value.  Such 
cycles  may  be  made  necessary  by  the  nature  of  the  medium  used. 
This  is  shown  in  Fig.  22.  1234  is  less  efficient  than  5678.  What 
is  meant  is  that  given  the  highest  and  lowest  possible  tempera- 
tures of  a  cycle,  the  greatest  efficiency  would  be  obtained  if  all 
of  the  heat  added  were  added  at  the  highest  temperature  and  all 
of  the  heat  removed  were  abstracted  at  the  lowest  temperature. 
Thus  in  Fig.  23  the  efficiency  of  the  steam  engine  with  saturated 

.    abed     ...     ,.          ...  .   aWc'd 

steam  is  — rj  while  that  with  superheated  steam  is  — rrr~  •     In 
eabf  eabb  g 

the  figure  — r^  is  practically  equal  to  — 7 —  and  by  adding  the 


FIG.  22. —  Cycle  with 
varying  temperature  on 
heat  line. 


e  f         9 

FIG.  23. — Steam  engine 
cycles  with  saturated  and 
superheated  steam. 


triangle  bb'h  to  numerator  and  denominator  the  expression  for 
the  superheated  cycle  is  obtained.  This  addition  having  a 
greater  effect  on  the  numerator  increases  the  efficiency.  Hence 
the  use  of  superheated  steam,  although  the  heat  is  not  added  at 
constant  temperature,  does  increase  the  efficiency  a  slight  amount. 
For  other  reasons  than  those  mentioned  here  the  use  of  super- 
heated steam  increases  the  efficiency  of  the  engine.  The  effi- 
ciency would  be  increased  by  a  greater  amount  if  the  heat  could 
have  been  added  on  the  dotted  line  d'V.  In  this  case  the  effi- 
ciency would  have  been 

d'b'c'd 

ed'b'g 


HEAT  ENGINES  AND  EFFICIENCIES  67 

In  some  machines  as  the  steam  turbine  the  indicated  work 
cannot  be  obtained  and  in  such  a  case  the  product  r/5  X  r}6  only 
can  be  determined. 


METHOD  OF  REPORTING  PERFORMANCE  OF  ENGINES 

Although  the  efficiency  of  an  engine  tells  an  exact  story,  for 
commercial  reasons  it  is  quite  common  to  report  the  performance 
of  a  heat  engine  in  pounds  or  cubic  feet  of  substance  required 
per  indicated  or  brake  horse-power  hour.  Thus  pounds  of  steam 
per  horse-power  hour  for  an  engine  or  turbine,  cubic  feet  of 
standard  gas  per  horse-power  hour  for  a  gas  engine,  coal  per  horse- 
power hour  for  a  gas  or  steam  engine  have  all  been  found  in  re- 
ports of  tests.  This  data  on  the  horse-power  hour  or  kilowatt- 
hour  basis  is  valuable  for  commercial  reasons  but  because  differ- 
ent gases  and  coals  have  different  heating  values,  and  steam  at 
different  pressures  and  qualities  contains  different  amounts  of 
heat,  these  statements  are  not  definite  until  other  data  are  known. 

When  coal  is  used  in  a  boiler  or  producer  the  coal  per  unit  of 
output  depends  on  the  efficiency  of  the  boiler  or  producer  as  well 
as  upon  the  engine.  To  separate  the  losses  and  find  out  just 
where  losses  occur  and  just  what  they  are,  it  is  always  better  to 
give  the  efficiency  of  the  heat  engine  as  a  percentage  using  the 
heat  supplied  as  the  base.  These  other  methods  are  valuable 
for  commercial  purposes  and  the  quantities  should  be  reported. 

A  method  used  in  reporting  the  performance  of  pumps  is  by 
duty.  This  is  the  amount  of  useful  work  in  foot-pounds  per  (a) 
100  Ibs.  of  coal,  (6)  1000  Ibs.  of  dry  steam,  or  (c)  per  1,000,000 
B.t.u.  The  first  two  methods  are  not  definite  although  the  third 
method  is  exact.  The  duty  by  the  third  method,  divided  by 
778,000,000,  will  give  the  overall  efficiency. 

Results  of  a  number  of  tests  will  now  be  given: 

HIGH-SPEED  NON-  CONDENSING  ENGINE 

Lh.p  ............   130  347.4-213 

77l~347.4+460~16-7% 


1190-1078.5+;*™(29.8-15)12.58 
'  1190-181.3 


=  14.5%         MK" 
Steam  pressure.  .  .    115.3  Ibs.  per  14'5     QA  7c/ 

^2  —  1  S  7  —  OQ.  /   /o 

sq.  in.  gauge  1D-' 


68  HEAT  ENGINEERING 

Barometer..          .    14.7  Ibs.   per  2546 

sq.  in.  *'~  30. 

Quality  of  steam..    1.00  8.3 


sq.  in.  '~  30.5[1190-  181.3J  " 


Pressure  at  end  of  expan-  _  120  _ 

sion.  .  .    15  Ibs.  per  sq.  in.  gauge  130 

Back  pressure. ...   0.3  Ib.  per  sq. 

in.  gauge 
Steam  per  i.  h.p.  hr.  30.5  Ibs. 

42.42 
B.t.u.  per  i.h.p.-min.  =  Q  ~gg  =  510 

HIGH-SPEED  NON- CONDENSING  COMPOUND  ENGINE 

I.h.p 130  in  =  16.7% 

B.h.p 115  773  =  14.5% 

Steam  pressure 115.3  Ibs.  per  sq.    772  =  86.7% 

in.  gauge 

Barometer 14.7 Ibs.  per  sq.  in. 2546  _117C7 

775     21.5  X[l  190- 181.3] 

Back  pressure 0.3  Ibs.  per  sq.  in.         _  it?m.g|  m 

gauge  14-5 

Pressure  at  end  of  ex-  115     cc  _ 

iciu  •  77  e  =  757,  =  88. 5% 

pansion...    15  Ibs.  per  sq.  in.  gauge 

Quality  of  steam ....    1.00 
Steam  per  i.  h.p.-hr. .  21.5  Ibs. 

42  42 
B.t.u.  per  i.h.p.-min.  =  -j^r=  362 

LOCOMOBILE  ENGINE 

I.h.p..  .    191  391+282-139 


Kw.  generator.  .       .  .  121.5  1352-1002 

773  "1352-  107   = 
Steam  pressure  .....   208  Ibs.  per  sq.  in.  (Complete  expansion.) 

gauge 
Feed  Temperature  ...  132°  F.  28.  1 


Barometer  .  .  .    1  4.  7  Ibs.  per  sq  .  in.  _  2546 

= 


_ 
775  =  9.9[  1352-  107] 

Vacuum  ...........    1  1.9  Ibs.  per  sq.  in.  20.6     ,_„.- 

774  =  287l  =  7'3-4% 

Degrees  superheat.  .   282>  F.  ^ 

Steam  peri.h.p.-hr.  .  9.9  Ibs.    Eff.  of  boiler 


Steam  per  Ib.  coal  .  .  .  8.28  Ibs. 

Heat  of  coal  ........    14,099  B.t.u.  per  Ib. 

42  42 
B.t.u.  per  i.h.p.-min.  =      on** 


HEAT  ENGINES  AND  EFFICIENCIES  69 

PUMPING  ENGINE 

.  861.34 


_.  „ 

771  ~  384.2+460"' 

839'8  llS9-926+^|(4-1.2)67.8 

173  =  — 


1189-76.0 

=  26.7% 
Steam  pressure 190.8  Ibs.  per   sq.  26.7 

??2  —  QO  17  =  °^  /O 

in.  gauge  o^./ 

Barometer 14.8  Ibs.  per  sq.  in. 2546          _ 

176 "  10.3711189-  76]  - 
Back  pressure. .        .1.2  Ibs.  per  sq.  in.  22.1 

abs.  7,4  =  2^ 

Pressure   at   end   of  839.8 

. ,,  .       ,        7/6  =  QA1  o=97.4% 

expansion 4  Ibs.  persq.  in.  abs.  001.0 

Quality  of  steam 0.99 

Steam  per  i.h.p.-hr .  .   10.37 

B.t.u.  per  i.  h.p.-min.  =  n  991 =  *92 

Duty  per  million  B.t.u.  =778,000,000X0.221  X0.974 
=  167, 000,000  ft.-lbs. 

STEAM  TURBINE 

Kw 6257  549-76 

771  ~559  +  460~ 

Steam  pressure 203.7  Ibs.  per  sq.  in.  1290-879 

abs.  ^  1290-  44  =33% 

Superheat..  .    165.5°  F.  33 

^  =  ^Q  =  72'5% 

Barometer..  .   29.92"  2546 

176  ~  11.95  X [1290-  44]  xO.746 

=  22.9% 

Backpressure 0.44  Ibs.  per  sq.  in.         _22-9 

,  ^4—  DO 

abs.  ***' 

Steam  per  kw.-hr ...    11.95  Ib. 

42  42 
B.t.u.  per  kw.-min.  =^746^0".229=249 

B.t.u.  per  elec.  h.p.-min.  =249X0.746  =  186 

PRODUCER  GAS  ENGINE 

I.h.p 579  2546 

775  ~  0.738  X  0.805  X  14320 
=  30% 

B.h.p 483  483 

776  =  579=83>5% 
Coal  per  i.h.p.-hr...  0.805 

Heating  value  of  coal  14,320  B.t.u.  per  Ib. 


70  HEAT  ENGINEERING 

Kind  of  coal  ........   Bituminous 

Efficiency  of  producer.  73.8% 

42.42 
B.t.u.  per  i.h.p.-min.  =       '       =  141.4 


141.4 
B.t.u.  per  b.h.p.-min.  =  Q~O^K  =  169 

BLAST-FURNACE  GAS  ENGINE 

I-h.p  ...................   775  2546 

775  =   110.5 
B.h.p  ..................   565  _5?5_ 

776  775  ~ 
Cu.  ft.  of  gas  per  i.h.p.-hr.  90.5 

Heat  value  of  gas  .......    110.5  B.t.u.  per  cu.  ft. 

42.42 
B.t.u.  per  i.h.p.-min.  =  Q~^KK  =  166 

1  AA 

B.t.u.  per  b.h.p.-min.  =  7^5    =  227 

U.  /  o 

DIESEL  OIL  ENGINE 

I-h-P  .............   523  2546 

775  ~  0.37  X  19270  ~ 
B.h.p  ............   450  450 

776  =  523  =  86% 
Oil  per  i.h.p.-hr.  .   0.37  Ib. 

Heat  value  of  oil  .  .    19,270  B.t.u.  per  Ib. 

42.42 
B.t.u.  per  i.h.p.-min.  =  ^  OKQ  =  H8 

118 

B.t.u.  per  b.h.p.-min.  =  —  —  =  143 
O.oo 

RESULTS  OF  VARIOUS  TESTS 


Engine  tested 

110-h.p.  Nurnberg  gas  engine  on  coke  

B.t.u, 
per  i.h.p- 
min. 

110 

B.t.u. 
per  b.h.p- 
min. 

138 

110-h.p.  Nurnberg  gas  engine  on  anthracite  coal.  . 
210-h.p.  Guldner  engine  on  illuminating  gas 

120 
100 

150 

600-h.p.  Ehrhardt  engine  on  coke-oven  gas  
11,000-kw.  Westinghouse  turbine  

113 

136 
213 

11,000-kw.  Curtis  turbine  
1,500-h.p.  locomotive 

350 

201 

10,000-h.p  marine  engine 

246 

2,200-h.p.  Corliss  engine  

226 

1.000-h.D.  air  -compressor  eneine.  . 

169 

TOPICS 

Topic  1. — What  is  a  heat  engine?     What  is  a  cycle?     What  is  the  general 
expression  for  the  efficiency  of  any  cycle?     What  is  the  expression  for  the 


HEAT  ENGINES  AND  EFFICIENCIES  71 

efficiency  of  the  Carnot  cycle?  What  does  this  efficiency  represent?  Give 
the  meaning  of  the  terms:  Carnot  efficiency,  type  efficiency,  theoretical 
efficiency,  practical  efficiency,  actual  thermal  efficiency,  mechanical  effi- 
ciency and  overall  efficiency. 

Topic  2. — Give  the  meaning  of  the  symbols:  771,  772, 773,  774,  775, 776,  rj.  Give 
the  relations  between  these  and  the  formulae  by  which  each  is  found.  Tell 
the  manner  of  determining  the  quantities  entering  into  these  formulae. 

Topic  3. — Give  the  conditions  for  maximum  efficiency  in  a  heat  engine  and 
show  that  although  these  conditions  may  not  be  fulfilled  high  efficiencies 
may  be  obtained.  Are  these  high  efficiencies  as  high  as  they  would  be  were 
the  conditions  for  maximum  efficiency  fulfilled?  How  are  results  of  tests 
reported? 

PROBLEMS 

Problem  1. — An  engine  using  35  Ibs.  of  steam  per  h.p.-hr.  at  125  Ibs.  gauge 
pressure,  x  =  0.98,  and  with  a  back  pressure  of  2.5  Ibs.  gauge,  has  its  con- 
sumption reduced  to  30  Ibs.  when  supplied  with  steam  under  the  same 
pressure  but  superheated  245°  F.  The  back  pressure  does  not  change. 
The  barometer  is  29.8  in.  What  is  the  per  cent,  saving,  if  any? 

Problem  2. — A  Corliss  engine  gives  the  following  test  results:  i.h.p.  = 
135.6;  b.h.p.  =  126.0;  steam  per  hour,  3406  Ibs.;  pressure  at  throttle  valve 
by  gauge,  135.6  Ibs.  per  square  inch;  barometer,  14.68  Ibs.  per  square  inch; 
pressure  at  end  of  expansion  by  gauge,  20.5  Ibs.  per  square  inch;  back  pres- 
sure by  gauge,  1.5  Ibs.  per  square  inch.  Find  the  various  efficiencies,  77,  776, 
775  and  771. 

Problem  3. — A  Corliss  condensing  engine  using  steam  at  174  Ibs.  gauge 
pressure  with  x  =  0.995  and  a  vacuum  of  27  in.  with  a  barometer  of  29.8  in. 
consumes  17.5  Ibs.  of  steam  per  kw.-hr.  output  from  generator.  Find 
overall  efficiency.  By  installing  a  low-pressure  turbine  and  by  reducing 
the  vacuum  to  28.5  in.  and  superheating  the  steam  to  200°  F.  of  superheat 
this  consumption  is  made  13.8  Ibs.  of  steam  per  kw.-hr.  and  the  output  of 
the  plant  has  been  increased  75  per  cent.  Is  the  change  of  value?  Why? 

Problem  4. — A  pumping  engine  gives  a  delivered  duty  of  175,000,000  ft.- 
Ibs.  per  1,000,000  B.t.u.  when  supplied  with  steam  at  165  Ib.  gauge  pressure 
with  x  =  1  and  a  temperature  of  hot  well  of  105°  F.  How  many  pounds  of 
steam  are  used  per  delivered  horse-power  hour?  If  the  mechanical  efficiency 
is  95  per  cent.,  what  is  the  steam  consumption  per  i.h.p.-hour? 

Problem  6. — A  gas  engine  has  an  overall  efficiency  of  24  per  cent.  The 
heat  lost  in  the  jacket  is  25  per  cent,  of  that  supplied  by  the  gas.  How 
much  heat  is  added  to  the  jacket  water  per  hour  if  the  delivered  power  is 
565  kw.? 

Problem  6. — The  results  of  the  test  of  a  producer  gas  engine  plant  are  as 
follows:  Coal  burned  in  120  hr.,  16,000  Ibs.;  heating  value  of  coal,  14,450 
B.t.u.;  gas  produced  during  test,  1,500,000  cu.  ft.;  heating  value  of  gas,  120 
B.t.u.  per  cubic  foot;  b.h.p.,  120;  i.h.p.,  150;  cooling  water,  1,000,000  Ibs.; 
temperature  of  inlet  water  65°  F.,  outlet  water  110°  F.  Find  the  efficiency 
of  the  producer.  Find  the  overall  efficiency,  77.  Find  the  indicated 
efficiency,  775.  Find  the  heat  removed  by  the  jacket  water.  Find  the  heat 
per  b.h.p. -hr.,  per  b.h.p.-min.,  per  i.h.p.-hour  and  per  i.h.p.-min. 


CHAPTER  III 
HEAT  TRANSMISSION 

The  phenomenon  of  the  transmission  of  heat  through  partitions 
is  very  important  as  it  enters  into  the  determination  of  the  areas 
of  radiators  for  heating  systems;  of  the  surface  required  in  con- 
densers, boilers,  evaporators,  intercoolers,  and  for  many  other 
engineering  structures.  Its  consideration  is  important  in  finding 
the  heat  required  for  warming  a  building  or  the  amount  of  re- 
frigeration to  keep  a  certain  cold  storage  warehouse  at  a  low 
temperature.  The  transmission  of  heat  is  complicated  and  al- 
though much  experimentation  has  been  done  the  exact  laws  are 
not  completely  determined.  The  following  discussion  is  based 
on  the  works  of  various  authors.  The  results  of  the  authors 
have  been  arranged  so  as  to  give  the  student  a  working  knowledge 
of  this  important  part  of  applied  thermodynamics. 

To  transmit  heat  from  one  body  to  another  or  from  one  part 
of  a  body  to  another  it  is  necessary  to  have  a  difference  of  tem- 
perature. Having  this,  the  heat  may  be  transmitted  by  one 
or  more  of  the  three  methods :  radiation,  convection  and  conduc- 
tion. Radiation  is  the  method  of  transmitting  heat  by  vibrations 
of  the  ether.  The  hot  body  starts  this  vibration  which  is  trans- 
mitted in  all  directions  through  the  ether  and  until  another  body 
receives  the  vibration  the  energy  is  not  sensible.  As  soon,  how- 
ever, as  a  body  opaque  to  these  vibrations  is  placed  in  the  path 
it  becomes  heated.  In  convection  heat  is  applied  to  some  mov- 
able body  and  then  this  energy  is  conveyed  by  the  actual  move- 
ment of  the  body  with  its  heat.  In  this  method  heat  is  carried 
by  moving  particles  of  matter.  Conduction  is  the  method  by 
which  heat  travels  from  the  hot  portions  of  a  body  to  another 
part  which  is  colder.  In  this  method  it  may  be  that  the  more 
violent  vibrations  of  the  particles  of  the  hot  portion  of  the  body 
are  gradually  transmitted  to  those  in  less  violent  vibration 
(colder)  or  according  to  the  later  views  of  the  constitution  of 
matter  it  may  be  that  more  electrons  are  thrown  off  at  higher 

72 


HEAT  TRANSMISSION  73 

temperature  vibrations  and  these  gradually  affect  the  vibrations 
of  the  atoms  at  the  more  remote  but  colder  portions  of  the  solid. 
The  laws  of  transmission  by  radiation  are  well  known  experi- 
mentally and  theoretically.  The  Stefan-Boltzmann  Law  for  the 
radiation  from  a  black  body  is 


Q  =  CF(2V  -  TV)  (1) 

Q    =  amount  of  heat  per  hour  in  B.t.u. 

C    =  a  constant  =  1.6  X  10~9. 

F    =  area  radiating  heat,  in  square  feet. 

TI  =  absolute  temperature  in  degrees  F.  of  hot  body. 

T%  =  absolute  temperature  in  degrees  F.  of  cold  body 
which  must  surround  the  hot  body  or  which 
must  include  all  rays  issuing  from  the  hot  body. 

It  is  to  be  remembered  that  the  formula  applies  to  black  bodies 
and  then  only  when  the  cold  surface  includes  all  rays  from  the 
hot  body.  If  the  surface  subtends  a  solid  angle  co  when  it  should 
have  subtended  a  solid  angle  of  2ir  or  a  hemisphere,  the  quantity 

Q  is  found  by  multiplying  (1)  by  —  • 

ZTT 

The  law  of  radiant  energy  has  been  under  discussion  from  the 
time  of  Newton  in  1690,  who  proposed  a  law  proportional  to  the 
first  power  of  the  temperature,  until  1879  when  Stefan  proposed 
that  the  law  be  of  the  form  given  in  (1)  basing  his  assumption 
on  some  experiments  of  Dulong  and  Petit,  Tyndall  and  others. 
After  some  years  (1884)  Boltzmann  proved  that  this  form  was 
correct  theoretically.  A  black  body  has  to  be  assumed  since  a 
colored  body  would  reflect  a  certain  amount  of  energy  and  thus 
the  amount  emanating  would  include  this  in  addition  to  the 
amount  radiated.  For  bodies  which  are  not  truly  black  (absorb 
all  radiation  falling  on  them)  the  law  is  approximately  true. 
The  nature  of  the  surface  has  also  an  effect  and  this  should  be 
taken  into  account.  Polished  bodies  will  not  radiate  as  much 
as  rough  surfaces. 

The  above  law  may  for  convenience  be  put  in  the  form 

(2) 


The  above  refers  to  a  black  body  of  perfect  radiating  power. 
The  relative  values  of  different  surfaces  by  Lucke  are  given  in  the 
following  table: 


74 


HEAT  ENGINEERING 


Absorbing 
power 

Porous  carbon 1 .  (X) 

Glass 0.90 

Polished  cast  iron 0 . 25 

Polished  wrought  iron 0 . 23 

Polished  steel 0 . 19 

Polished  brass 0 . 07 

Hammered  copper 0 . 07 

Polished  silver 0. 03 

The  transmission  of  heat  by  pure  conduction  is  a  simple 
matter.  The  law  for  this  is  similar  to  that  of  the  transmission 
of  electric  current.  The  flow  of  heat  is  proportional  to  the  differ- 
ence of  temperature  and  the  area  of  the  substance  and  is  inversely 
proportional  to  the  length  of  path  or  thickness.  Thus 


(3) 


Q  =  B.t.u.  per  hour. 
F  =  area  in  square  feet. 
I    =  thickness  in  feet. 
ti  —  temperature  at  one  side  in  degrees  F. 
tz  =  temperature  at  one  side  in  degrees  F. 
C  =  coefficient  of   conduction   in   B.t.u.    per   hour 
per  square  foot  per  degree  F.  for  1  ft.  thickness. 


FIG.  24.— 
Transmis- 
sion surface. 


Values  of  C 

C  =  C0[l  +  a(t  -  32)] 
t  =  temperature  in  degrees  F. 


Substance 

Co 

a 

Substance 

Co 

or 

Air  

0.03  to  0,012 

0.0011 

Cork  powdered 

0.03 

Brass             .  .  . 

53.5 

0.0011 

Glass  

0.54 

Brick  

0.46 

Iron  

48.5 

-0.00012 

Carbon  

0.8 

Limestone  

1.35 

Carbon  dioxide 

0.006 

Masonry  

0.46 

Cast  iron  

40.5 

-0.00012 

Sandstone  

0.87 

Concrete  

0.46 

Steel  soft  

26.7 

Copper  

239.0 

0.00003 

Water  

0.292 

Cork  board  .  .  . 

0.17 

Wood  

0.10 

HEAT  TRANSMISSION  75 

The  value  of  C  for  gases  has  been  shown  by  Maxwell1  to  be 
given  by 

C  =  Kr,cv 

K  =  constant  between  0.5  and  2.5. 
77  =  coefficient  of  viscosity  =  the  force  between  two 
planes  separated  by  distance  unity  when  one  plane 
is  moving  with  unit  velocity  relative  to  other. 

cv  =  specific  heat  at  constant  volume. 

He  also  showed  that  C  varies  as  the  %  power  of  the  absolute 
temperature. 

Nusselt  examined  insulating  materials  and  found  that  the 
conductivity  of  these  substances  increased  as  the  absolute 
temperature. 

The  amount  of  heat  carried  by  convection  depends  on  the 
amount  of  substance  involved,  its  specific  heat  and  its  tempera- 
ture. This  is  simple  to  find  but  the  amount  of  heat  which  can 
be  abstracted  from  these  particles  by  a  given  surface  under  given 
conditions  is  a  complex  matter  and  it  is  this  method  which  will 
be  considered  throughout  this  chapter.  This  transfer  of  heat 
through  partitions  is  the  important  one  in  most  applications  of 
heat. 

When  heat  is  transmitted  through  a  surface  as  in  a  boiler  or  as 
in  an  indirect  heating  coil  there  must  be  films  of  substance  on  each 
side  of  the  wall  which  prevent  the  passage  of  heat.  Suppose, 
for  instance,  in  a  boiler  where  the  temperature  of  the  hot  gases 
on  one  side  is  1500°  F.  and  the  temperature  of  the  water  on  the 
other  is  325°  F.,  the  thickness  of  the  steel  tube  is  Y±  in.  and  that 
there  are  4  Ibs.  of  water  evaporated  per  hour  per  square  foot 
under  these  conditions. 

The  heat  transmitted  per  square  foot  per  hour 

=  4  X  r325 

=  4  X  889.8  =  3559.2. 

For  the  steel  wall,  by  the  law  of  conduction,  there  results: 

26  7 

3559'2  -  X  (<1  ~  '2) 


''  -  <*  =  =  2'78 

il.  Mag.,  1860. 


76  HEAT  ENGINEERING 

Now  the  difference  between  the  water  and  the  gas  is  1175°  F. 
and  of  this  there  is  only  a  drop  of  2.78°  F.  in  the  steel  wall.  A 
thin  scale  of  Jioo  m-  on  the  water  side  and  ^{Q  in.  of  soot  on  the 
gas  side  would  account  for  considerable  drop. 

For  the  scale,  C  will  be  taken  as  1. 

3559.2 


12000 
For  the  soot,  C  will  be  taken  as  0.1. 


' 


16  X  12  X  0.1 


This  accounts  for  188°  of  the  1175°,  showing  that  there  must  be 
still  further  resistance  at  the  surface.  This  of  course  must  be 
due  to  films  of  water  and  of  gas.  It  is  seen  from  the  tables  of 
values  of  C  that  water  has  about  thirty  times  the  value  of  C  for 
air  and  gas,  and  hence  there  will  probably  be  J^i  of  987°  drop  in 
the  water  film  and  3%i  of  987°  drop  in  the  gas  film  if  these  are  the 
same  thickness.  The  gas  is  probably  less  thick  and  if  the  water 
film  is  taken  as  three  times  the  thickness  of  the  gas  film  the  drops 
in  each  will  be  approximately  90°  in  the  water  and  897°  in  the  gas 
film. 

The  thicknesses  to  give  these  results  are  found  below  assuming 
C  for  water  0.292  and  0.009  for  the  gases.  This  is  the  mean  of 
0.006  and  0.012. 

X    12  =  0.089  in. 


X  12  =  0.0272  in. 

Thus  it  is  seen  that  a  film  of  small  thickness  on  either  side 
accounts  for  a  great  drop  which  must  exist  to  explain  the  low 
conductive  powers  of  the  heating  surface.  If  it  were  not  for 
the  films  and  scale  the  heat  transmitted  by  the  steel  tube  would 
be 

26  7 
Q  =  -^-(1500  -  325)  =  1,505,000  B.t.u.  per  hour. 

This  would  mean  an  evaporation  of  1690  Ibs.  of  water  per 
square  foot  of  heating  surface.  Fig.  25  shows  the  temperature 
gradient  across  these  surfaces.  It  appears  from  the  figure  and 
the  calculation  above  that  anything  which  would  tend  to  decrease 


HEAT  TRANSMISSION 


77 


T 


the  thickness  of  the  films  of  water  or  gas  would  cut  down  the 
drop  of  temperature  in  these,  putting  a  greater  drop  in  the  wall 
and  so  causing  an  increase  in  the  heat  transmitted.  One  method 
of  doing  this  is  to  increase  the  velocity  of  the  substance  bringing 
heat  up  to  the  surface  or  taking  heat  from  the  surface.  This  has 
been  tried  and  found  true.  The  increase  of  velocity  of  the  gas 
or  water  wipes  some  of  the  film  away  decreasing  its  thickness. 
This  simple  explanation  is  evident  and  shows  why  one  would 
expect  an  increase  in  the  heat  transmitted,  if  either  the  velocity 
of  the  gas  or  the  velocity  of  the  water 
were  increased  across  the  surface 
transmitting  the  heat. 

In  October,  1909,  Prof.  W.  E.  Dalby 
presented  to  the  Institute  of  Mechan- 
ical Engineers  of  Great  Britain  a  paper 
on  Heat  Transmission,  "the  purpose 
of  which  was  to  place  before  the  mem- 
bers of  the  Institution  a  general  view 
of  the  work  which  has  been  done  re- 
lating to  the  transmission  of  heat 
across  boiler- heating  surfaces."  He 
has  given  this  view  and  with  it  a  list 
of  406  references  to  various  papers 
dealing  with  this  subject.  They  are 
listed  chronologically,  by  Authors  and 
by  Subjects,  and  the  student  is  re- 
ferred to  this  paper  for  most  of  the  lit- 
erature on  this  important  subject. 

The  fact  that  the  velocity  of  the 
gases  affected  the  rate  of  transmission 
was  experimentally  known  as  early  as  1848  but  there  does  not 
seem  to  be  a  statement  of  this  until  1874  when  Prof.  Osborne 
Reynolds  read  a  paper  before  the  Literary  and  Philosophical 
Society  of  Manchester  (Vol.  xiv,  1874,  p.  9)1  "On  the  Extent 
and  Action  of  the  Heating  Surface  for  Steam  Boilers."  In  this 
very  short  paper  of  a  few  pages,  Reynolds  points  out  the  im- 
portance of  the  subject.  He  states  that  the  heat  carried  off 
from  the  gas  is  proportional  to  the  internal  diffusion  of  the  fluid 
at  or  near  the  surface,  that  is,  on  the  rate  at  which  the  particles 


FIG. 


25. — Temperature 
gradient. 


1  See  also  The  Steam  Engine,  by  John  Perry,  p.  594. 


78  HEAT  ENGINEERING 

which  give  up  their  heat  diffuse  back  into  the  hot  gas.     This  ac- 
cording to  him  depends  on  two  things: 

1.  The  natural  internal  friction. 

2.  The  eddies  caused  by  visible  motion  which  mix  up  the 
fluid  and  continually  bring  fresh  particles  into  contact 
with  the  surface. 

The  first  of  these  is  independent  of  the  velocity,  and  the  second 
term  is  dependent  on  the  velocity  and  density  of  the  fluid, 
the  heat  transmitted  at  any  point  being 

Q  =  At  +  Bmwt  (4) 

where         Q  =  heat  per  square  foot  per  hour 
A,  B  =  constants 

t  =  difference  of  temperature 
m  =  density  of  substance,  pounds  per  cubic  foot 
w  =  velocity  along  surface  in  feet  per  second. 

He  then  discusses  the  quantities  A  and  B  and  the  way  in  which 
the  heating  surface  may  be  of  more  value  by  an  increase  in 
velocity.  He  finally  closes  his  paper  with  the  hope  of  giving  a 
further  communication.  He  does  not  give  the  values  of  A  and  B. 

In  Reynolds'  paper  he  gave  no  theoretical  discussion  nor  deri- 
vation of  his  formula. 

In  1898,  Professor  John  Perry  gave  the  following  theoretical 
discussion  to  prove  that  transmission  varied  as  the  velocity: 

In  a  thin  film  the  number  of  molecules,  n,  entering  the  film  per 
square  foot  of  area  is  equal  to  the  number  leaving  and  the  friction 
force  developed  by  the  axial  momentum  given  up  by  the  n  mole- 
cules as  they  are  arrested  from  velocity  w  will  be  given  by 

P(per  square  foot)  oc  nw  (5) 

Now  since  the  kinetic  energy  is  proportional  to  the  tem- 
perature the  change  in  the  energy  as  the  surface  abstracts  heat 
will  be 

Q  oc  n(t  -  0)  (6) 

t  =  original  temperature  of  gas 
6  =  temperature  at  surface  of  pipe. 

Eliminate  n  from  (5)  and  (6) 

)  (7) 


HEAT  TRANSMISSION  79 

It  is  known  that  the  force  of  friction  expressed  in  feet  head 
when  a  fluid  flows  through  a  pipe  is*  proportional  to  the  square 
of  the  velocity  w*.  To  change  this  to  pounds  per  square  foot  it 
must  be  multiplied  by  the  weight  of  1  cu.  ft.,  m,  or 

P  oc  mw2  (8) 

Hence  (7)  reduces  to 

Q  =  K'mw(t  -  0)  (9) 

where  Kf  =  constant  of  equality. 

If  now,  Perry  points  out,  there  is  a  film  of  gas  at  the  side  of 
the  surface  of  thickness  b  and  conductivity  K,  there  will  really 
be  a  drop  from  t'  to  9  in  the  film  and  only  a  drop  from  t  to  t' 
in  the  gas  so  that 

Q  =  K'mw(t  -  t')  =  y  (f  -  0) 
*  jfomtf  +      * 


or  Q  =  K'mw 


t  - 


--  +  K'mw 


K'mwt  +  ~0 


+  K'mw 


K'mw[t  -  B] 


1  + 


bK'mw 


(10) 


K' 

be  called  K"  the  formula  will  be  the  same  as 


oK  mw 
~K~ 

(9)  with  a  different  coefficient.  If  b  decreases  as  w  increases, 
which  is  probable,  the  product  bw  will  be  nearly  constant  and 
since  K  is  large  and  m  is  small  for  most  gases  the  denominator 
of  the  fraction  will  change  little.  Hence  formula  (9)  is  practically 
correct. 

This  equation  of  Perry  and  that  of  Reynolds  may  be  changed 
by  remembering  that 

M  =  mFw 

where     M  =  total  weight  per  second  in  pounds 
F  =  area  of  passage  in  square  feet 
w  =  velocity  in  feet  per  second 
m  =  weight  per  cubic  foot. 

M 

Hence  -=-  =  mw 

r 


80  HEAT  ENGINEERING 

M 
and  Q  =  At  +  B  -= 


Q  =  K"(t  -  0)  (100 


In  Perry's  formula  the  fact  that  P  varied  with  m  and  w2 
might  have  been  made  to  include  the  fact  that  P  also  varies 
inversely  with  the  mean  hydraulic  depth,  d\.  Thus: 


The  hydraulic  depth  or  hydraulic  radius  is  equal  to  the  area 
of  the  cross  section  of  a  passage  carrying  a  fluid  divided  by  the 
perimeter  of  the  cross  section. 


Hence  Q  =  K'         (t  -  0)  (11) 

ai 

This  formula  states  that  the  smaller  the  tube  the  greater  the 
heat  transmission. 

Dalby  shows  that  in  1888  Ser  and  in  1897  Mollier  gave  for- 
mulae in  which  the  heat  transmitted  depended  on  the  square 
root  of  the  velocity  of  the  gases  along  the  tube,  while  Werner, 
Halliday,  Carcanagues  and  Brille  show  that  it  depends  on  the 
velocity.  In  1897  T.  E.  Stanton  investigated  the  "Passage  of 
Heat  between  Metal  Surf  aces  and  Liquids  in  Contact  with  Them" 
(Phil.  Trans.  Roy-Soc.,  Vol.  cxc  A,  p.  67)  and  showed  that  the 
transmission  of  heat  varied  with  the  velocity  of  water,  and  that 
although  not  stated  in  the  paper  in  words,  the  formula  of  Rey- 
nolds is  applicable  to  the  liquid  side  as  well  as  the  gaseous  side  of 
a  surface. 

Prof.  John  T.  Nicolson  (Junior  Institution  of  Engineers, 
Jan.  14,  1909,  and  London  Engineering,  Feb.  5,  1909,  p.  194) 
and  H.  P.  Jordan  (Institution  of  Mech.  Eng.  of  Great  Britain, 
Dec.  1909,  p.  1317)  have  each  proposed  formulae  for  the 
transmission  of  heat  which  are  somewhat  similar  to  that  of  Rey- 
nolds, but  in  these  as  in  the  works  of  Stanton  the  coefficient  of 
transmission  depended  on  the  temperature  difference. 

In  general,  the  formula  for  the  transmission  of  heat  through 
a  surface  is 

Q  = 


HEAT  TRANSMISSION  81 

where     Q  =  B.t.u.  transmitted  per  hour 

ti  =  high  temperature  of  fluid  on  one  side  in  degrees  F. 
t2  =  low  temperature  of  fluid  on  other  side  in  degrees  F. 
F  =  surface  in  square  feet 

K  =  coefficient  of  transmission  in  B.t.u.  per  hour  per 
square  foot  per  degree  F. 

According  to  Reynolds 

K  =  A  +  By,  or  A  +  Bmw  (11') 

•  According  to  Perry 

K  =  K",  or  K"mw  (11") 


or  later  K  =  K'"  (11"') 


Finally  it  has  been  shown  experimentally  that  K  varies  with 
the  temperature  difference.  In  the  above  three  values  of  K  it 
is  seen  that  for  a  given  tube  with  a  given  discharge  the  value 

M 

of  K  would  be  constant,  since  mw  =  -^-. 

r 

This  part  does  not  depend  on  the  variation  of  temperature 
along  the  pipe.  The  quantities  A,  B,  K"  and  K'"  may  however 
depend  on  temperature.  In  general  the  temperature  along  a 
surface  changes  from  point  to  point  so  that  t\  —  £2  is  a  varying 
quantity  and  to  apply  the  values  of  K  determined  by  experiment 
it  will  be  necessary  to  find  what  is  the  mean  difference  in  tem- 
perature if  the  temperatures  of  the  substance  at  the  ends  of  the 
surface  are  given.  The  formula  to  be  used  for  heat  transfer  is 
then 

Q  =  (K  for  mean  AQ(Mean  At)F  (12) 

MEAN  TEMPERATURE  DIFFERENCE 

The  question  then  arises:  What  is  the  value  of  mean  A£? 
There  is  evidence  to  show  that  K  =  , ..  •  To  determine  an  ex- 
pression for  mean  M  for  any  value  of  n,  two  cases  will  have  to  be 
considered,  one  for  any  value  of  n  except  zero  and  the  other  for 

n  =  0.     The  reason  for  this  is  the  fact  that  I  xn'dx  =    ,_,    ^  xn' +1 
for  all  values  of  n'  except  —  1.     For  nf  =  —  1  the  value  of  the 


82  HEAT  ENGINEERING 

integral   becomes  log  x.     To  find   the  values  of  mean  AZ  the 
two  cases  are  considered. 
First  case: 

K  =  constant,  or  n  =  0 

dQ  =  K(thx  -  tcx)dF  =  -  3600  Mhchdth  =  ±  36QQMcccdtc      (13) 
dQ  =  heat  per  hour  in  B.t.u.  for  surface  of  dF 

sq.  ft. 
K  =  heat  per  hour  in  B.t.u.  per  square  foot  per 

degree 

t'hx  —  temperature  in  degrees  F.  in  warm  sub- 
stance at  point  x 

tcx  =  temperature  in  degrees  F.  in  cool  sub- 
stance at  point  x 

Mc  and  Mh  =  weight  of  substances  flowing  per  second 
cc  and  ch  =  specific  heat  of  substances  in  B.t.u.  per 
pound  per  degree  F. 

Temperature  of 
""^ — L^Cool  Substance 

_      ^  __£ 
Transmission- 


Temperature  of~ 
Warm  Substance 

FIG.  26.— Parallel  flow. 

The  minus  sign  before  Mh  is  used  because  dth  is  negative  as 
dF  increases,  measuring  from  the  inlet  end  of  the  warm  sub- 
stance, where  the  temperature  is  tiA,  toward  the  outlet  where  the 
temperature  is  fa. 

If  the  warm  and  cool  substances  flow  in  the  same  direction  the 
arrangement  is  called  a  parallel  flow  arrangement  (Fig.  26)  and 
the  temperature  of  the  cold  substance  increases  with  dF  hence 
dtc  is  positive.  With  the  warm  substance  flowing  in  the  opposite 
direction  from  the  flow  of  the  cool  substance  the  arrangement  is 
called  counter  flow  (Fig.  27).  A  minus  sign  is  required  before 
the  dtc  as  the  temperature  decreases  with  an  increase  of  dF. 

If  tic  is  the  temperature  of  the  cool  substance  at  the  end  cor- 
responding to  the  entrance  of  the  warm  substance  and  t^c  is  that 
of  the  cool  substance  at  the  point  of  exit  of  the  warm  substance, 
the  following  is  true. 

Mhch(tlh  -  t2h)  =  +  Mccc(tlc  -  t2c)  (14) 


HEAT  TRANSMISSION 


83 


The  upper  (minus)  sign  is  for  parallel  flow  and  the  lower 
(plus)  sign  is  for  counter  flow  in  this  equation.  This  of  course 
assumes  that  all  the  heat  leaving  the  warm  substance  goes  into 
the  cool  substance  and  hence  there  is  no  radiation  loss. 

The  following  notation  may  be  used: 


and  from  (13)  it  is  seen  that 

JlA    ~   txh)    =    +    McCc(tlc   -   txc) 

—     /        ^     _l        J. 

Vi  isxh)    "T~   ^lc 


=    txk    -   txc    =    txh    ± 


•*•  MeCe  txh  +   Mccc 


Temperature  of 
Cool  Substance 


Cool  ^^. 
Substance 


Substance 


Temperature  of 

Warm  Suhstauce 


FIG.  27.—  Counter  current  flow. 


Now 


(15) 


1  ± 


Mhch 


-  Mccc 
Substituting  this  in  (13)  gives 


3600 


Mccc 


dF  =  - 


F  =  - 


d\tx 


K  Mhch 

1  ± 


Mccc 


3600 
K 


McCc 


(16) 


(17) 


84 


HEAT  ENGINEERING 

H  =  X(mean  At)F  =  3QWMhch(ti 

360o;i 
meanA£  = 


Now 


Substituting  for  .F  its  value  from  (17) 

Mhch(ti  - 


meanA£  = 


, 

log 


(18) 


From  (14) 


tic- 


MCCC  tih  —  tzh 

tih   —   tic   —    (t%h 


Equation  (18)  reduces  to 
mean  At  = 


tih  — 


-  At* 


(180 


(19) 


This  is  independent  of  the  direction  of  flow.     The  form  of  the 
expression  is  the  same  for  parallel  or  counter  current  flow.     It 


Counter  Current  Parallel  Current 

tc  Outlet  .>  t h  Outlet  tc  Outlet  <th  Outlet 

FIG.  28. — Temperature  range  for  various  arrangements 


Constant  Warm  Temperature 
Outlet  <th    Inlet 


will  be  found  that  for  given  temperature  ranges  the  value  of 
mean  A£  will  be  greater  for  counter  current  flow.  If  TH  is  the  same 
for  all  of  the  surface,  T&  and  Tzh  are  the  same,  and  are  so  used 
in  finding  the  A^s.  If  the  lower  temperature,  as  in  the  case  of  a 
boiler,  is  constant  this  formula  takes  care  of  that  condition. 
If,  however,  both  temperatures  are  constant  as  in  the  case  of  an 
evaporator  then  this  formula  reduces  to  %  which  of  course  has 
the  true  value  Th  —  Tc.  The  temperature  ranges  are  shown 
in  Fig.  28. 


HEAT  TRANSMISSION  85 

With  this  value  of  mean  AZ  determined  by  the  differences  of 
temperature  the  value  of  Q  for  a  given  F  could  be  determined  as 
soon  as  K  is  known  or  F  could  be  found  for  a  given  Q. 

Q  =  K  X  mean  A«  X  F  (9) 

This  of  course  is  true  for  the  case  in  which  K  does  not  vary 
with  A£x.  The  equation  could  be  used  to  determine  K  if  Q,  F, 
and  mean  A£  are  measured. 

Kf 

Second  case:  K  =  TTTT^ 

(AtzJ 

~K' 

dQ  =  7-— r-n  AWF  =  -  360O3f ACfcdks  =  ±  3600  Mcccdtxc     (20) 

(/Uz; 

By  the  method  used  in  reducing  (17)  the  following  results 
3600Af*c* 


(22) 


(mean  AQ 

(mean 

Substituting  from  (22)  and  using  (18r) 

f  . 

(meanA<>'" 


Mkch 


t\h  — 


(24) 

This  equation  gives  the  value  of  Q  or  F  if  the  other  is  known  in 
terms  of  Kr  which  is  not  the  transmission  per  degree  per  square 
foot  per  hour,  but  the  constant  which  when  multiplied  by 
(mean  A£)~n  gives  the  coefficient  of  transmission.  The  value  of 
Kf  in  (24)  as  in  (12)  may  vary  with  any  other  quantities  than 
temperature  and  if  the  value  of  K'  for  the  conditions  of  the  given 


86 


HEAT  ENGINEERING 


problem  as  to  velocity,  hydraulic  radius  or  density  be  substi- 
tuted the  expression  will  be  correct. 

It  is  well  to  notice  at  this  point  that  — *—^ — -  gives  the  value 


of  mean  A£  within  J^  of  1  per  cent,  for  any  value  of  n  if  A^ 
is  less  than  Ko  °f  ~^~o — 2*      The  formulae 


A*  2 


or 


mean  At  = 


mean  A£  = 


need  not  be  employed  to  find  mean  AZ  if  the  change  in  A£  is  slight. 
If  however  Ati  —  A£2  is  large  compared  with  the  mean  AZ  then  the 
formulae  must  be  used.  Having  mean  A2  for  any  given  problem 
the  heat  or  surface  may  be  found  by 


or 


Q  =  K'  X  (mean  A*)1-"  X  F 
Q  =  K  X  mean  AZ  X  F 


depending  on  whether  or  not  K  varies  with 
_ey   the  temperature. 

DETERMINATION  OF  K 

The  method  of  finding  this  K  will  now  be 
considered. 

FIG.  29. — Trans-  In  Fig.  29  let  the  various  temperatures  at 
mission  through  tne  division  lmes  between  the  various  layers 
various  layers.  .  J 

be  ti ,  t  ,  ^2     and  tz    while  ^i  represents  the 

temperature  of  the  gas  and  t2  represents  the  temperature  of  the 
water.     The  following  results  for  1  sq.  ft.  area: 

Q  =  K(ti  -  h) 


Cln 


HEAT  TRANSMISSION  87 

where  c1     =  coefficient  of  heat  conduction  for  gas  film. 

c11    =  coefficient  of  heat  conduction  for  soot. 
c111  =  coefficient  of  heat  conduction  for  heating 

surface. 

clv    =  coefficient  of  heat  conduction  for  scale. 
cv     =  coefficient  of  heat  conduction  for  water  film. 

r,  I11,  P1,  I™,  f  are  the  thicknesses  in  feet  of  the  gas  film,  soot, 
heating  surface,  scale,  and  water  film  respectively. 

Now  Q  ^  =  Zi  -  ii1 

7n 

Q-  i  l        i  n 
~  *i    -  <i 


71V 

~ 


r/i  7ii          /in          7iv  7v  1 

Adding  these  «£  +  L+  J_  +  ^  +  1_J  _  fl  _ 


Q  =  p  --  pi  -  i          F     ^  («i  -  «  (25) 

^T+  c"  ~^~  c1"      c1^      c^" 

but  Q  =  K[ti  -  t2] 

Hence  K  (for  combination)  =  T  --  ^  -  ^  -  ^  -  ^r       (26) 

—  +  —  +  —  +—  +  — 

C1          C"         C111    '     CIV        Cv 

To  apply  this  formula  the  thickness  of  the  gas  film  I1  and  of 
the  water  film  lv  must  be  known.  These  of  course  depend  on  the 
velocity.  The  value  of  c1  for  the  gas  and  cv  for  the  water  depend 
on  the  temperature  of  these  and  probably  on  their  viscosity  and 
on  the  hydraulic  radius  of  the  pipe.  The  values  of  the  c's  for 
the  solid  materials  vary  with  the  temperature  and  hence  the  co- 
efficient of  conduction  K  for  the  combination  will  depend  on  the 
temperature. 

In  the  case  of  the  coefficient  of  transmission  for  walls  and 
partitions  of  buildings,  the  air  film  usually  depends  on  the 


88  HEAT  ENGINEERING 

exposure  and  kind  of  surface  hence  for  this  formula,  when  applied 

c1 
to  building  calculations,  the  terms  for  the  two  films,  =  (ti  —  t1  1) 

cv 
and  -^   (?z  —  £2),  are  replaced  by  ai(ti  —  £  i)  and  a2  (t\  —  ^2). 

The  quantities  a  are  given  by  empirical  formulae  which  depend 

Q 
on  the  variable  factors,    y  is  replaced  by  a.     K  becomes 


,          ,+-+- 
ai  /  c11  T  cm  T  civ  T  a2 

The  values  of  K  for  walls  and  partitions  in  most  building 
problems  are  used  as  independent  of  temperature  since  the  varia- 
tion in  temperature  is  not  great  in  these  cases. 

c1    cv 
To  determine  the  value  -^  ,^  ,  ai  or  a2  is  difficult  in  the  case  of 

transmission  from  gas  to  water,  as  in  the  case  of  boilers  or  inter- 
coolers,  from  water  to  water  in  the  case  of  sterilizers  or  from  water 
to  gas  as  in  the  case  of  indirect  hot-  water  heaters.  These  terms 
are  the  largest  and  most  effective  terms  entering  into  the  value 
of  K  and  for  that  reason  the  drop  of  temperature  in  the  plate 
has  been  assumed  to  be  zero.  The  coefficients  at  each  surface 
have  been  determined  and  this  determination  has  been  made 
experimentally,  putting  the  heat  into  the  form 

Q  =  Kg(h  -  6)  =  KW(S  -  t,)  (28) 

where  &  is  the  mean  temperature  of  the  heating  surface. 

Kg   is  the  coefficient  of  transmission  on  the  gas  side  = 

c1 

^  =  B.t.u.  per  square  foot  per  hour  per  degree. 

Kw  is  the  coefficient  of  transmission  on  the  water  side 


ti  =  temperature  of  gas,  tz  =  temperature  of  water. 

The  values  of  these  K's  may  depend  on  velocity,  temperature 
or  anything  else  but  they  are  determined  by  experiment. 

If  it  is  desired  to  eliminate  the  temperature  of  the  surface,  this 
may  be  done  by  finding  the  value  of  K  in 

Q  = 


HEAT  TRANSMISSION  89 


K  = 


L_  (29) 


for  which  1     ,   _1_ 

XV0          JL\.W 

This  assumes  that  the  drop  in  the  heating  surface  and  its  scale 
is  zero  or  is  included  in  the  drop  in  each  film. 

If  the  expression  for  K  is  assumed  to  be  that  of  Perry, 

M 
Kg  =  K"gmgwg  =  K"g^ 

F  a 

M 

T?  ~K~H     71/f    /,,«  If  w 

Kw  =  K  WMWWW  =  K  w-^r~ 

r  w 

These  are  the  same  as  the  values  of  Reynolds  in  which  A  has 
been  made  zero  and  B  has  been  made  K". 

Nicolson  suggests  that  3  be  used  for  K"  for  gases  and  6  be 
used  for  the  K"  for  water.  These  values  are  independent  of  the 
temperature  and  using  them  in  (29)  the  following  results: 


, 


K      QMg         Mw       3mgwg       Qmwww  (29') 

6~~       "  ~~ 


(30) 

mwww 


L_      ft  +  LV 

www       \    L    / 


_(_&L_\MW       /     6    \M, 
--~-: 


K  L    /  Qmwww       \    L    /  Qmgwg 

(31) 

In  (31)  or  (32)  K  is  expressed  in  terms  of  water  or  gas  condi- 
tions and  of  the  quantity  L  which  is  the  ratio  of  conditions  of  flow 
of  gas  to  that  of  water. 

>v\ 

It  will  be  remembered  that  for  a  perfect  gas  ra  =  j—f,  and  this 

£>  1 

may  be  used  to  find  m  for  any  gas. 


90  HEAT  ENGINEERING 

For  different  values  of  L  the  values  of  K  are  given  by  the  table 
below. 

L  K 

2.0  3.00  mwww  or  1.50  mgwg 

1.0  2.00  mwww  or  2.00  mgwg 

0.5  1.20  W^WM,  or  2.40  mgwg 

0.2  0.55  mwww  or  2.73  m^ 

0.1  0.29  mwww  or  2M  mgwg 

0.01  0.03  Wu,^  or  2.99  m0wg 

0.0  0.0  mwww  or  3.00  mgwg 

Since  air  or  flue  gas  at  580°  F.  weighs  about  J^5  Ib.  per  cubic 
foot  and  a  common  velocity  is  100  ft.  per  second  and  hot  water 
which  weighs  about  60  Ib.  per  cubic  foot  may  move  with  a  ve- 
locity of  12  ft.  per  second,  a  common  value  of  L  will  be  given  by 

X  100  _   J_ 
'          " 


"  mwi0w  "     60  X  12      '  180 

and  K  =  3.00  mgwg  (33) 

Since  X  is  assumed  independent  of  t, 

Q  =  3.00m,W,^^-2F  (34) 

!  ^-At-l 

10g'A^ 

Before  leaving  this  formula  it  may  be  well  to  point  out  that 
Nicolson  also  suggests  bringing  into  this  expression  the  different 
diameters  of  the  surface  in  contact  with  the  gas  and  the  water. 

Thus  dQ  =  Kgmgwg(ti  —  6}irdgdx  =  Kwmwww(d  —  t^irdwdx      (25) 

9-  *2  =   Kgmgwgdg    =  L, 
ti  —  e      Kwmwwwdw 

e==  iT^  +  rrL7'1 

*i  -  e  =  f(il  ~  W  (37) 


(ll  ~  ^  (38) 


If  Kgmgwg  is  known  it  may  be  used  for  the  determination  of 
the  heat,  of  L'  and  (ti  —  tz)  or  their  mean  values  are  known. 


HEAT  TRANSMISSION  91 

Thus  Q  =  KgmgWg  i  _L_  Lf  mean  (ti  —  t2)F 

This  can  be  done  for  the  water  side 
Hence  K  =  Kgmgwg  1   ,  L, 

T,       1  r  dg 
since  L   =  ~  L  -j- 

A         dw 

and  Kg  =  3  according  to  Nicolson. 

This  result  is  similar  to  equation  (31)  except  for  the  term  -r-. 


The  equations  (37)  and  (38)  enable  one  to  compute  the  heat 
transfer  from  either  the  water  or  the  gas  side. 

The  value  of  2.75  for  the  coefficient  of  (33)  is  recommended  by 
Nicolson. 

The  values  of  Kg  =  3  for  gas  have  been  determined  by  Prof. 
Nicolson  from  seven  sets  of  experiments  in  which  the  tempera- 
tures of  the  water  and  air  were  varied.  This  variation  however 
was  not  sufficient  to  show  the  variation  of  Kg  with  temperature. 
The  values  obtained  varied  from  2.84  to  3.11  but  Nicolson  did 
not  attempt  to  show  the  variation  of  this  with  temperature. 
The  value  of  Kw  =  6  was  determined  from  Stanton's  experiments 
in  which  water  only  was  used.  In  this  Stanton  points  out  that 
Kw  does  vary  with  the  temperature.  Nicolson  proceeds  further 
to  show  that  K  varies  with  temperature,  using  superheated 
steam  to  transmit  heat  to  air  or  water  and  using  heated  air  to 
transmit  heat  to  water.  In  addition  to  these  experiments 
Nicolson  discussed  certain  boiler  tests  to  determine  constants 
for  a  formula.  He  uses  the  formula 

Q  =  K(T  -  e)F  (39) 

and  shows  that 


(40) 

Q  =  B.t.u.  per  hour. 

T  =  mean  temperature  of  gas  along  flue  in  degrees  F. 
6  =  mean  temperature  of  "flue  in  degrees  F.  =  tem- 
perature of  steam. 

4>  =  \(T+6). 


92  HEAT  ENGINEERING 

di  =  hydraulic  mean  depth  of  flue  in  inches. 

area       _  diam. 
perimeter  ~~       4 

ma  =  weight  of  1  cu.  ft.  of  gas  in  pounds. 
wg  =  velocity  of  gas  in  feet  per  second. 
F  =  area  in  square  feet  of  surface  on  gas  side. 
M  =  total  weight  per  second. 
M 


This  formula  is  worked  out  from  experiment.  It  is  correct  in 
form  if  K  is  of  the  form 

K  =  AM  +  BM^. 
Q  will  be  of  the  form 

Q  =  F[AM&2  +  %B(Mi/2  +  A^)A^A*21/2] 

or  since  AZiAZ2  may  be  written  as  At2  and  %(A$iM  +  A^~)  may  be 
written  as  A^.  This  reduced  to 

Q  =  A[M  +  %BM*]MF 

This  is  the  form  of  (39)  with  the  value  of  K  substituted  when 
the  value  of  A£  is  inserted. 

These  are  the  same  and  hence  the  formula  is  correct  in  form. 
If  the  constants  are  worked  out  to  fit  certain  conditions,  it  will 
lead  to  correct  results  when  applied  to  such  conditions- 

H.  P.  Jordan  in  the  Transactions  of  the  Institution  of  Mechan- 
ical Engineers  for  Dec.,  1909,  p.  1317  gives  the  formula  in  the 
form 

Q  =  3600  i  0.0015  +  [o.  000506  -  0.00045di 

+  0.00000165  (^-£^)]fr)  (T  -  0)F    (41) 

This  is  also  seen  to  be  of  correct  form  as  is  the  case  of  Nicolson's 
formula.  For  the  range  of  the  experiments  the  results  are 
usable.  Unfortunately  these  two  formulae  do  not  yield  the 
same  results  for  various  problems.  For  instance,  using  a  problem 
given  by  Nicolson  the  differences  may  be  seen. 

In  a  Lancashire  boiler  the  flue  is  36  in.;  400  Ibs.  of  coal  are 
burned  per  hour  with  24  Ibs.  of  air  per  pound  of  coal.  The 
gases  leave  the  fire  at  2200°  F.  and  at  the  end  of  the  flue  they 
are  900°  F.  The  steam  temperature  is  350°  F. 


HEAT  TRANSMISSION  93 


0  =  350°  F.  assumed 


400  X  25  _ 
3600  X  7  = 


By  (40): 

i-qcn          i  -i 

K  =  IJgg  +  ^  V950  X  1.11  X  0.4J  =  4.75  +  0.34  =  5.09 


By  (41): 

K  =  3600  {0.0015  +  [0.000506  -  0.00045  X  9  + 

0.00000165(950)]0.4}  =  2.64 

Using  the  method  first  proposed  by  Nicolson  and  assuming  a 
velocity  of  3^  ft.  per  second  for  the  water, 


L  =         -  -  °-013 


K=  (2+^013)  a4  =  1- 


This  third  method  is  evidently  incorrect  due  to  the  temperature 
effect  and  there  is  a  great  discrepancy  between  the  results  from 
the  formula  of  Nicolson  and  that  of  Jordan. 

If  these  had  been  used  for  a  mean  temperature  of  950°  but 
with  4  in.  diameter 

By  (40):  K  =  4.75  +  0.61  =  5.36. 

By  (41):  K  =  7.5 

rp     i     n 

For  —  £  —  =  300  and  for  a  number  of  4-in.  pipes  of  same  area 
the  results  would  have  been  as  follows: 


94  HEAT  ENGINEERING 

300        1 
By  (40):  X  =  200  +  40  V30°  X  2  X  °'4 

=  1.5  +  0.346  =  1.8 
By  (41): 

K  =  3600  {0.0015  +  [0.000506  -  0.00045  X  1  + 

0.00000165  X  300]  04}. 

=  3600  [0.00172]  =  6.2 

There  is  no  change  in  the  third  method,  K  being  1.2  as  before. 
This  is  more  nearly  the  value  1.8  which  is  for  temperatures 
near  the  values  used  in  the  experiments.  The  Jordan  formula  is 
reduced  from  a  large  number  of  experiments  and  in  none  of 
them  does  K  fall  below  4.5. 

The  formula  of  Nicolson  does  not  agree  with  that  of  Jordan 
and  apparently  the  reason  for  this  may  be  said  to  be  due  to  the 
fact  that  Nicolson's  formula  has  been  derived  for  large  flues 
with  high  temperatures  and  Jordan's  has  been  derived  from  small 
flues  and  for  mean  temperatures  of  300  °.  Hence  it  would 
be  well  to  restrict  these  to  the  conditions  from  which  they  were 
derived.  The  simpler  formula  for  K  is  evidently  in  error  on 
account  of  the  effect  of  temperature.  The  wide  variations  make 
it  necessary  to  compare  the  results  to  be  obtained  from  other 
experiments. 

H.  Kreisinger  and  W.  T.  Ray  in  Bulletin  18  of  the  Bureau  of 
Mines,  U.  S.  Dept.  of  Interior,  on  the  Transmission  of  Heat  into 
Steam  Boilers,  have  shown  that  the  quantity  of  heat  from  hot 
gas  to  hot  water  varies  directly  with  the  velocity  and  with  the 
temperature  to  some  power.  The  efficiency  of  the  heat  trans- 
mission increases  with  the  decrease  of  diameter  of  tube. 

Wilhelm  Nusselt  (Mit.  liber  Forschungsarbeiten,  Heft  89) 
in  his  article  on  the  transfer  of  heat  in  tubes  has  reduced  the 
theoretical  form  for  K  with  an  empirical  constant  for  the  value 
of  K  in  the  formula 

A  4  \4 

Q  =  KF 


} 

logeAT2 


0-786 

(42) 


U-  \       A     / 

in  greater  calories  per  hour  per  square  meter  per  degree  C. 


HEAT  TRANSMISSION  95 

coefficient  of  conduction  of  gases  at  wall 
temperature,  greater  calories  per  square 
meter  per  hour  per  degree  for  1  meter 
thickness. 

w  =  velocity  in  meters  per  second. 

Cp  =  specific  heat  at  constant  pressure  for  1 

cubic  meter  at  condition  of  gas  in  flue. 
X  =  coef.  of  conduction  of  gases  at  mean  tem- 
perature of  tube. 
d  =  diam.  of  tube  in  meters. 

This  may  be  thrown  into  a  different  form  since 


C1      —    r        ^ 

^P  -  CP  BT 
cp  =  specific  heat  of  1  kg.  of  gas. 


(42)  then  becomes 

X  /inrtr  \  °-786 

*^^fcM3B)  ^ 

To  change  this  to  K  in  B.t.u.  per  square  foot  per  hour  per  degree 
the  same  factor  15.90  is  used  when  the  terms  have  the  following 
meaning  : 

\vaii  =  coef.  of  conduction  of  gas  at  wall  tempera- 
ture of  tube  in  B.t.u.  per  hour  per  square 
foot  per  degree  F  for  1  ft.  thickness. 
w  =  velocity  in  feet  per  second. 
Cp  =  specific  heat  at  constant  pressure  for  1  cu. 

ft.  of  gas  under  conditions  in  tube. 
X  =  .coef.  of  conduction  of  gas  at  temp,  in  tube. 
d  =  diameter  in  feet. 

The  values  of  X  for  the  different  gases  are  given  below,  quoted 
from  Nusselt: 

French  English 

Air  0 . 01894(1  +  228  X  10~50.  0 . 01287[1  +  127  X  10~5(Z  -  32)] 
CO2  0 . 01213(1  +  385  X  W~5t).  0 . 00814[1  +  215  X  10~5(£  -  32)] 
Steam  0.0192  (1  +434  X  10~50.  0.01288U  +  241  X  KT5(f  -  32)] 

Illuminating  Gas 

0.0506  (1  +  300  X  10~50.  0.03390[1  + 167  X  10~5  (t-  32)] 
For  a  4-in.  pipe  with  1500°  F.  gas  and  a  wall  at  400°  F.  with  a 


96  HEAT  ENGINEERING 

velocity  of  50  ft.  per  second,  and  assuming  gas  to  be  a  mean  be- 
tween air  and  CO2,  the  following  results: 

X  =  0.01050[1  +  171  X  10~5(1500  _  32)]  =  0.0371 
0.01050U  +  171  X  10~5(400   _  32)]  =  0.0171 


CP  =  0.24  X 

.   0.0171  /50  X  0.0049\  °-786 
^V      0.0371      / 


=  15.90  X  0.0171  X  1.265  X  5.55 
=  1.91 

For  a  36-in.  flue  for  the  mean  temperature  of  1550°  F.  and  a 
wall  temperature  of  350°  F.  with  mgwg  =  0.4  the  following 
results : 

WgWlgCp     =     WgCp 

Xutau  =  0.0105[1  +  171  X  10~5  X  318]  =  0.0162 
=  0.0105[1  +  171  X  10~6  X  1518]  =  0.0377 

v  -      K  on  Q-0162  /0.24  X  0.4\  °-786 
L5.90   30>214  \    0.0377    j 

,15.90X0.0162 
1.266 

For  a  4-in.  pipe  for  1550°  F.  for  gas  and  350°  F.  for  wall  with 
wgmg  =  0.4,  the  result  is 

K  =  i  *  on     °-0162  /°-24  X  0.4\  °-78' 
(M)°'214V    0.0377    / 

=  15.90  X  0.0162  X  1.266  X  2.2  =  0.716 

These  do  not  check  with  results  of  Nicolson  nor  Jordan  for 
high  temperatures,  and  of  course,  since  based  on  experiments 
in  which  the  temperature  did  not  rise  to  such  high  values  it 
should  not  be  used  in  such  cases.  The  formula  has  been  com- 
puted for  values  of  A£  of  about  20°  C.  to  80°  C. 

In  1914,  F.  E.  McMullen  performed  a  large  number  of  experi- 
ments in  the  Mechanical  Engineering  Laboratory  of  the  Rensselaer 
Polytechnic  Institute  on  the  transmission  of  heat  from  hot  air 
to  water  through  brass  and  steel  pipes.  The  velocity  of  the  water 
was  varied  from  1  to  8  ft.  per  second,  that  of  the  air  from  2  to  17 
ft.  per  second.  The  brass  pipes  were  of  %-in.  and  %-in.  outside 
diameter  and  the  steel  pipe  was  of  %-m.  diameter.  The  mean 


HEAT  TRANSMISSION  97 

temperature  difference  was  as  much  as  300°  F.  The  results  of 
this  series  of  experiments  showed  that  K  varied  inversely  as  A£^ 
and  directly  with  the  density  of  the  air  and  the  velocities  of  the 
water  and  gas.  The  formula  becomes 

Q  =  £(mean  At)F  (43) 

126m(wq-  1.75)°-  W* 
* 


or  Q  =  K'(meanAO%F 

K'  =  126m(wa  -  1.75)°'W*  (45) 

m  =  weight  of  1  cu.   ft.    of   air   at   mean   temperature  in 

pounds  =  -—• 

wa  =  velocity  of  air  in  feet  per  second. 
ww  =  velocity  of  water  in  feet  per  second. 

(meanAO  = 
A 


From  the  above  discussion  of  problems  it  is  suggested  that 
Nicolson's  second  formula  be  used  for  boiler  problems;  that 
Jordan's,  Nusselt's  and  the  Rensselaer  formula  be  used  for  air 
coolers  and  heaters  where  mean  AZ  is  about  30|00  F. ' 

STEAM  CONDENSERS 

For  transmission  of  heat  through  condenser  tubes  a  reference  is 
made  to  the  work  of  G.  A.  Orrok,  A.  S.  M.  E.  Transactions, 
Vol.  xxxii,  p.  1138.  In  this  he  shows  that  the  constants  of  heat 
transmission  from  the  steam  to  the  water  depends  on  the  tem- 
perature difference  to  the  J^  power,  on  the  square  root  of  the 
velocity  of  the  water,  on  the  square  of  the  steam  richness,  on  the 
material  and  its  cleanness.  In  his  formula 


Q  =  #(mean  M)F  =  ^"(mean  ti*F  (47) 

_  JTapVV^  f.  . 

=  (meanAO* 


98  HEAT  ENGINEERING 

Q  =  B.t.u.  per  hour. 

F  =  square  feet  of  surface. 
K  =  B.t.u.  per  square  foot  per  hour  per  degree. 
K'  =  630. 

a  =  cleanness  factor,  1  to  0.5. 

p  =  steam    richness    factor  =  —f  where  ps  is  the   pressure 

corresponding  to  temperature  of  steam  in  condenser 
and  pt  =  actual  pressure  in  condenser  measured 
in  any  units. 

ww  =  velocity  of  water  in  feet  per  second. 
M  =  materials  factor  =  1.00  for  copper,  0.98  for  admiralty 
metal,   0.87    for  aluminum    bronze,   0.80     for    cupro 
nickel,  0.79  for  tin,  0.75  for   monel   metal,   and  0.74 
for  Shelby  steel. 

The  student  is  also  referred  to  Orrok's  article  for  a  bibliography 
of  heat  transmission  for  condensers. 

AMMONIA  CONDENSERS 

In  regard  to  ammonia  condensers  this  formula  should  apply, 
but  experimental  results  reported  in  the  Transactions  of  the 
American  Society  of  Refrigerating  Engineers  for  1907  seem  to 
indicate  a  much  lower  result.  This  is  given  by  a  curve  in  which 
K  for  double-pipe  condensers  is  given  by 

K  =  130\A^  (50) 

ww  =  velocity  of  water  in  feet  per  second 
and  Q  =  ^(mean  At)F 

(mean  At)  —  — - — 


BRINE  COILS 

For  brine  cooling  in  double  pipes  the  value  of  K  seems  to  be 
given  by 

K  =  84wb 

EVAPORATORS  AND  FEED-WATER  HEATERS 

The  formula  of  Orrok  should  be  applicable  to  the  heating  and 
boiling   of  water  by  steam  as  in  evaporators  and  feed-water 


HEAT  TRANSMISSION  99 

heaters.  The  constants  and  values  have  been  determined  for 
various  steam  pressures  and  for  that  reason  they  should  be  ap- 
plicable. However  this  may  be,  two  formulae  recommended  by 
Hausbrand  will  also  be  mentioned. 

One  of  the  formulae  quoted  is  that  due  to  Mollier  from  Hage- 
mann's  experiments  reduced  to  English  units : 

K  =  10  +  1 110  +  0.6  (t8  +  ^-^-2)  }  VW  (51) 

t,  =  temperature  of  steam  in  degrees  F. 

t\  =  temperature  of  liquid  at  entrance  in  degrees  F. 

tz  —  temperature  of  liquid  at  exit  in  degrees  F. 

The  formula  which  Hausbrand  recommends  for  the  transmis- 
sion of  heat  from  steam  to  water  which  is  boiling  is  that  due 
to  Jelinek's  experiments.  When  reduced  to  English  units  it 
becomes 


d  =  diameter  of  tube  in  feet. 
I  =  length  of  tube  in  feet. 

In  the  above  formula  the  material  is  copper  and  steam  is  on 
the  inside.  This  formula  may  be  correct  for  such  apparatus  in 
which  there  is  a  low  velocity  of  the  liquid.  This  is  the  only 
reason  for  such  a  formula  applying,  as  all  experiments  show 
that  the  heat  varies  with  the  velocity  of  the  liquid.  For  wrought 
iron  0.75  of  the  above  values  are  to  be  used,  0.5  for  cast  iron  and 
0.45  for  lead.  In  evaporators  with  dilute  solutions  of  15  per 
cent,  solid  matter  in  solution  the  transmission  is  decreased  by 
about  15  per  cent,  of  the  above  for  clean  water  while,  with  thick 
viscous  liquids,  K  is  about  one-third  of  that  given  above. 

HEAT  FROM  LIQUIDS  TO  LIQUIDS 

The  coefficient  of  transmission  from  a  liquid  to  a  liquid  may  be 
found  from  the  formula  (29')  given  by  Nicolson  in  which  the 
constant  6  is  used  in  each  term. 

-•      —          —          —  (S3) 

K      Qmwww      Qmwww      3mwww  *     ' 


100  HEAT  ENGINEERING 

Hausbrand  gives  the  following  equation  for  the  coefficient 
from  one  liquid  to  another  through  brass  and  copper  walls. 

K  m  _  300 

JL      .  +  .     _J_  (54) 


This  refers  to  greater  calories  per  square  meter  per  degree  C  for 
velocities  in  meters  per  second  and  was  calculated  by  Mollier 
from  Joule's  researches.  Hausbrand  suggests  that  66  per  cent. 
of  the  value  of  K  be  used  for  practical  purposes.  To  change  this 
to  English  units  the  following  calculations  are  made  where  Kf 
refers  to  French  units  and  Ke  to  English: 


v12 

Wf   =     We   X 


K 


(55) 


These  two  formulae  do  not  give  the  same  results  and  the 
later  one  is  recommended  for  general  use. 

FACTOR  OF  SAFETY 

In  all  of  the  above  expressions  for  transmission  it  would  be  well 
to  reduce  the  quantity  by  %  to  have  an  excess  of  surface,  or  what 
would  be  the  same  thing  increase  the  surface  by  33  per  cent,  in 
finding  the  area  required  in  a  problem. 

RADIATORS 

The  coefficient  of  transmission  for  direct  radiators  for  heating 
varies  with  conditions  of  the  surface  and  the  type  of  radiators. 


HEAT  TRANSMISSION  101 

Carpenter  quotes  tests  varying  from  1.23  to  1.97  while  Rietschel 
gives  values  of  K  varying  from  0.51  to  2.65.  A  value  of  K 
of  1.8  will  be  used  as  an  average  for  the  computation  of  the 
heat  transmitted  for  all  types  of  direct  radiators.  This  gives 
for  such 

Q  =  1.8(f.  -  ta)F  (56) 

Q  —  B.t.u.  per  hour. 

F  =  square  feet  of  surface. 

ts  =  temperature  of  steam  or  water  in  degrees  F. 

ta  =  temperature  of  room  in  degrees  F. 

For  indirect  heaters  the  formula 


may  be  used  in  which 

K  =  2  +  l.75\/wa. 

wa  =  velocity  of  air  across  surface  in  feet  per  second. 

ti  =  temperature  of  air  at  inlet.  in  degrees  F. 

tz  =  temperature  of  air  at  outlet  in  degrees  F.. 

ts  =  temperature  of  steam  in  degrees  F. 

Experiments  have  been  made  and  plotted  for  the  determination 
of  the  transmission  of  heat  and  the  resultant  air  temperatures 
obtained  with  coils  and  cast-iron  radiators  and  these  are  to 
be  resorted  to  rather  than  formulae  because  the  resultant  tem- 
perature is  dependent  on  the  velocity  and  the  temperature  of 
the  entering  air  and  steam.  Since  in  most  problems  low  pres- 
sure steam  is  used  the  curves  constructed  from  data  of  the 
American  Radiator  Co.  and  of  the  Buffalo  Forge  Co.  are  given, 
the  first  for  a  cast-iron  pin  radiator  known  as  a  Vento  Heater 
and  the  second  for  the  sections  of  pipe  coils,  each  section  being 
four  pipes  deep.  The  curves  of  Figs.  30  and  32  give  the  result- 
ant air  temperatures  when  air  at  zero  degrees  enters  the  indirect 
heater  and  passes  over  one,  two,  three,  four,  five,  six,  seven  or 
eight  sections  at  different  velocities.  The  velocities  are  found  by 
finding  the  volume  of  the  air  at  70°  F.  and  then  computing  the 
velocity  by  dividing  this  volume  by  the  clear  area  through 
which  the  air  passes.  Thus  the  velocity  to  be  used  for  the 
curve  is  not  the  actual  velocity  at  entrance  or  exit  but  is  that 
occurring  when  the  temperature  of  the  air  is  70°  F. 

The  curves  of  Figs.  31  and  33  are  those  giving  the  average  B.t.u. 


102 


HEAT  ENGINEERING 


per  square  or  lineal  foot  of  the  total  heater  under  the  conditions 
given.     It  will  be  seen  that  as  the  velocity  increases  the  final 


200 


400 


1400 


1600    1800 


600    800    1000    1200 
Velocity  in  Ft.  per  Minute 

FIG.  30. — Temperature  of  air  leaving  Vento  Heater  with  air  at  0°  F.  and  veloc- 
ity at  70°  F.     Steam  at  5  Ibs.  gauge  pressure. 

2700 


200 


400 


1400 


1600 


1800 


600          800  1000         1200 

Velocity  in  Ft.  per  Minute 

FIG.  31. — Heat  transmitted  per  sq.  ft.  of  Vento  Heater  with  5  Ibs.  steam  and 
air  entering  at  0°  F.     Velocity  at  70°  F. 

temperature  of  the  air  falls  but  that  the  heat  transmitted  per 
square  foot  increases,  due  to  the  lower  final  temperature  and  the 


HEAT  TRANSMISSION 


103 


200 


400 


1400 


1600 


1800 


600  800  1000  1200 

Velocity  inJFt.  per  Minute 

FIG.  32. — Temperature  of  outlet  from  sectional  heaters  of  four  coils 
each,  with  air  entering  at  0°  F.     Steam  at  5  Ibs.  gauge  pressure.    Velocity 


1200 
1100 
1000 
900 
800 
700 
600 
500 
400 
300 
200 
100 


5.9 


200 


400 


1800 


500    800    1000    1200    1400    1600 
Telocity  in  Ft.  per  Minute 

FIG.  33. — Heat  transmitted  per  lineal  foot  of  1  in.  pipe  for  4  row  sections 
of^oil  heaters  with  5  Ibs.  steam  and  air  entering  at  0°  F.     Velocity 'at 


104  HEAT  ENGINEERING 

higher  velocity.  As  the  number  of  sections  is  increased  the  final 
temperature  rises  by  decreasing  increments  since  the  tempera- 
ture difference  between  the  steam  and  air  is  less  for  the  succes- 
sive sections,  decreasing  the  heat  transmitted,  and  the  heat  per 
square  foot  is  less  for  the  same  reason. 

Since  air  often  is  supplied  at  a  higher  temperature  than  0°  F. 
it  is  well  to  understand  how  such  curves  may  be  used  for  these 
conditions.  Suppose  air  at  40°  F.  and  600  ft.  velocity  is  to  be 
heated  in  Vento  heaters.  It  will  be  seen  that  one  section  would 
have  been  required  to  give  this  temperature  and  if  a  temperature 
of  100°  F.  had  been  desired  three  sections  would  be  required  to  do 
this  with  zero  air.  The  first  section  would  bring  the  air  to 
40°  F.  so  that  only  two  sections  will  be  required  for  40°  air. 
Now  with  zero  air  one  section  at  600  ft.  velocity  will  transmit 
1450  B.t.u.  per  hour  per  square  foot  and  three  sections  will 
transmit  1150  B.t.u.  per  square  foot  per  hour.  The  amount 
transmitted  by  the  last  two  sections  will  therefore  be 

1150  X  3  -  1450  =  2000 
or  2     =  1000  B.t.u.  per  square  foot. 

If  two  sections  had  0°  F.  air  supplied  the  curves  show  that 
1275  B.t.u.  per  hour  per  square  foot  would  have  been  trans- 
mitted but  in  this  case,  the  entering  air  being  at  40°  F.,  there 
is  a  smaller  A£  and  hence  a  smaller  transmission  per  square  foot 
per  hour. 

These  curves  can  be  used  with  certainty  as  they  are  the 
results  of  experiment  and  they  displace  any  formula  for  compu- 
tations. They  are  better  in  this  case  as  they  show  what  may 
be  obtained  actually  in  practice. 

HEAT  THROUGH  WALLS  AND  PARTITIONS 

The  heat  transmitted  through  walls  and  partitions  is  com- 
puted by 

Q  =  K(t,  -  t*)F  (58) 


where  I       V_       V_  1  (59) 

ai  *  c'  T  c"  ~          '  a2 

Q  =  B.t.u.  per  hour. 

ti  =  temperature  on  one  side  in  degrees  F. 
£2  =  temperature  on  other  side  in  degrees  F. 
F  =  area  in  square  feet. 


HEAT  TRANSMISSION  105 

The  value  of  a  is  found  by  experiment  in  the  form 


d  =  constant  depending  on  condition  of  air. 

=  0.82  air  at  rest  in  rooms  or  channels. 

=  1.03  air  with  slow  motions,  as  over  windows. 

=  1.23  air  with  quick  motions,  as  outside  of  build- 

ing. 

e  =  constant  depending  on  material. 
T  =  temperature  difference  of  film  of  air  at  wall. 

Values  of  e 

Cast  iron  ......................................  0  .  65 

Cotton  and  fabrics  ..............................  0.  65 

Charcoal  .......................................  0.  71 

Glass  ..........................................  0.60 

Metal,  polished  .................................  0  .  05 

Masonry  .......................................  0  .  74 

Paper  .........................................  0  .  78 

Rusted  iron  ....................................  0  .  69 

Water  .........................................  1.07 

Wetted  glass  ...................................  1  .09 

Wood  .........................................  0.74 

For  masonry  walls,  of  thickness  I  ft. 

T  =  16.2  -  4.00Z  (61) 

For  wood  T  =  1.8°  F  (62) 

For  glass  T  =  ^fa  -  t2)  (63) 

The  values  of  c  are  given  below: 

Air  (still)  .....................................  0.03 

Brass  .........................................  61  .00 

Building  paper  ................................  0  .  08 

Cork  .........................................  0.17 

Cotton  and  felt  ................................  0.02 

Glass  ............  .............................  0.54 

Masonry  and  plaster  ...........................  0.46 

Sandstone  ....................................  0  .  87 

Sawdust  ......................................  0  .  03 

Slate    ........................................  0.19 

Terra  cotta  ....................................  0.54 

Tin  ..........................................  35.60 

Wood..  0.12 


106  HEAT  ENGINEERING 

Experiments  show  that  the  values  of  c  change  with  the 
temperature,  but  for  the  temperatures  used  in  practice  the 
variation  is  not  great. 

EFFICIENCY  OF  HEAT  TRANSMISSIONS 

The  efficiency  of  a  heating  surface  is  the  ratio  of  the  heat 
abstracted  by  the  surface  from  the  gases  to  the  amount  of  heat 
which  the  gas  could  give  up  to  the  surface.  If  TV  is  the  tem- 
perature of  the  hot  substance  entering  and  T^h  is  the  temperature 
of  the  hot  substance  leaving  the  tube  and  Tc  is  the  lowest  tem- 
perature of  the  cool  substance  in  contact  with  the  opposite  side 
of  the  surface,  the  expression  for  efficiency  is 

_  Tih  —  Tzh  _  1        Tzh  —  Tc  _  -•  _  A^2  ,,*.<. 

"   Tlh-Tc~~        ~  Tlh  -Tc~       ~  Ah 

This  is  true  since  the  heat  available  in  M  Ibs.  of  gas  of  specific 
heat  c  is 

Qi  =  Mc(Tlh  -  Te) 

and  the  amount  utilized  is 

Qi  -  Q2  =  Mc(Tlh  -  T2h) 

In  a  given  experiment  the  temperatures  are  known  but  with  a 
given  T^  and  Tc  and  with  a  given  surface  it  is  necessary  to  find 
Tzh  in  order  to  ascertain  the  efficiency. 

There  are  two  cases  to  consider: 

First,  if  the  cool  substance  changes  in  temperature  along  the 
surface. 

Second,  if  the  cool  substance  remains  at  a  fixed  temperature  as 
in  the  case  of  a  boiler. 

In  the  first  case  it  has  been  shown  earlier  in  the  chapter  that 

-Mhchdth  =  K(txh  -  txc)dF 

Mhch         1          , 
—  ab  =  —  ~—  -r\  —      t  \  dth 

&       (txh   —   txc) 

MhCh  1        dAt  dAt 

TT     ~M^~M   =const-  KKt 

1  *  Mccc 
If  K  is  independent  of  temperature  this  becomes 

I  +  MhCh 


w  ccc  i 

FK  -  jfij—    -  =  loge  Ty- 


HEAT  TRANSMISSION  107 

Jlc   —   J2c  Jlfe   —   Jlc   —    (tzh  —   J2c) 


_ 

1    ±  =  " 


ZTZT  1  ~         2  l  ~  , 

Hence      FlL  ,,      ,.  --  -T  =  -     —  -  =  loge  T- 


jT*  (66) 

This  shows  that  the  efficiency  of  a  surface  increases  with  the 
increase  of  area,  with  the  increase  of  the  constant  K  or  of  the 
difference  of  Ai,  but  decreases  as  Qi  increases.  If,  however,  the 
area  F  is  increased  for  a  fixed  length  of  tube  by  increasing  the 
diameter,  this  apparently  shows  that  the  efficiency  will  increase 
with  the  diameter.  It  is  known,  however,  that  K  varies  with  the 
velocity  and  hence  this  term  would  decrease  with  the  square  of  the 
diameter  so  that  the  product  FK  would  decrease  with  the  increase 
of  diameter  and  hence  the  efficiency  would  decrease.  On  the 
other  hand,  if  d  is  decreased  for  a  given  length,  the  term  FK  is 
increased.  In  the  above  discussion  Qi  and  (Aii  —  Ai2)  have  been 
assumed  to  be  constant  but  this  would  not  be  so  for  if  d  were  de- 
creased the  velocity  would  be  increased  and  with  it  Qi  for  a  given 
length  would  be  increased,  but  (Aii  —  Ai2)  would  increase  to  a 
greater  degree  and  the  final  result  would  be  an  increase  of 
efficiency. 

If  K  depends  on  (Ai)~n,  the  integration  gives 

•1   r  .   .  .    „! 


K'F 


Qi  n 

nFK'        Aii"  -  Ai2n 


«1 

nFK' 

Aii  —  Ai2 
/A*,          \ 

U(,  •  v 

Qi     /                 Ai2          _t  \ 

Eff.  =  77  =  1  -  ^Ep —^  (67) 

Ai7  " 

In  this  as  before  the  efficiency  increases  with  FK'  and  de- 
creases with  Qi. 


108  HEAT  ENGINEERING 

In  the  second  case  the  temperature  is  constant  on  the  cold  side 
and  the  expression  then  becomes 

+  Mhchdth  =  K(txh  -  tc)dF 

+  if  dt  is  measured  from  cold  to  hot. 
If  K  is  independent  of  the  temperature  this  becomes 

KdF          dth 

Mhch  ~  txh  —  tc 

Tlh  -  Tc 


F    =  i^eT2h-  Tc 
Tih-Tc          **L 

7F, TFT  =   £   Mhch 

-iZh  —  lc 

KF 
77    =  1    —   €      M™ 

In  this  as  before  the  efficiency  increases  with  the  product  KF 
and  decreases  with  the  quantity  M.  Thus  if  the  diameter  of  a 
pipe  is  increased  the  velocity  decreases  for  the  same  M  or  M 
increases  for  the  same  velocity.  In  each  of  these  the  exponent  is 
decreased  since  M  would  vary  with  the  square  of  d  directly  and 
K  inversely.  Hence  the  large  diameter  means  a  smaller  effi- 
ciency. If  M  is  increased  for  a  given  tube  by  increasing  the  ve- 
locity K  will  increase  in  the  same  ratio  and  the  efficiency  may  not 
be  changed.  If  the  length  is  increased  F  is  greater  and  the 
efficiency  increases.  Since  this  increase  of  efficiency  is  logarith- 
mic, the  increase  beyond  a  certain  value  of  F  is  very  slow. 

If  K  depends  on  the  temperature  the  following  results 

(69) 


This  leads  to  an  expression  similar  to  the  one  used  previously. 

Kreisinger  and  Ray,  in  Bulletin  18  of  the  Bureau  of  Mines, 
show  by  results  of  tests  on  an  experimental  boiler  that  the  effi- 
ciency decreased  with  the  velocity  to  a  certain  point  although  the 
quantity  of  heat  increased.  In  these  experiments  the  value  of 
M  increased  more  rapidly  than  the  coefficient  of  conduction. 
Of  course  if  at  the  same  time  F  were  increased  to  its  proper 
amount  the  efficiency  would  increase.  In  these  experiments  it 
was  shown  that  the  absorption  of  heat  increased  with  the  in- 
crease of  the  initial  temperature  although  the  rate  of  increase 


HEAT  TRANSMISSION  109 

of  heat  gradually  diminished.  The  enlargement  of  the  di- 
ameter of  the  tubes  reduced  the  efficiency.  Increasing  the 
length  of  a  tube  increased  the  efficiency  of  the  surface. 

For  a  boiler  Kreisinger  and  Ray  have  used  the  form  for  the 
value  of  K  suggested  by  Perry. 

K  =  K'mw 
Hence  Mhchdt  =  K' 

Mh 
but  mhwh  =  -~r- 

r  o 

dt    K' 


K' 


r,  =  1  -  e 


K'FX 

Foe,,  (70) 


This  expression  for  efficiency  does  not  change  with  change  in 
weights  and  therefore  does  not  give  curves  similar  to  those  found 
by  experiment. 

If  the  Reynolds  form  be  used 


Mhchdth  =  K'(A  +  Br 
this  leads  to  (A+B£)  Fx 


and 


This  equation  yields  a  curve  similar  to  those  found  experi- 
mentally. 

PROBLEMS 

A  number  of  problems  will  now  be  computed,  applying  the 
formulae  best  suited  for  the  work. 

Problem  1.  —  A  4-in.  boiler  tube  is  used  in  a  return  tubular  boiler  with  a 
grate  of  six  times  the  flue  area.     Fifteen  pounds  of  coal  are  burned  per 


110  HEAT  ENGINEERING 

square  foot  of  grate  per  hour  with  25  Ibs.  of  air  per  pound  of  coal.  The  gases 
entering  the  tubes  are  1350°  F.  and  at  exit  they  have  been  reduced  to  600°  F. 
What  is  the  average  value  of  the  heating  surface  of  tubes  and  how  many 
pounds  of  water  at  the  boiling  point  will  be  evaporated  if  the  pressure  is 
130.3  Ibs.  gauge? 

The  formula  to  use  in  this  case  is  that  of  Nicolson  (40). 

The  gases  are  of  practically  the  same  density  as  air,  hence  B  =  53.35. 
7r4      6  X  15  X  (25+  1)      nnKOfr,, 

Mass  of  gas  per  tube  =  yrrX —     — ^60O —     — ==  "-0567  Ibs.  per  second. 

1350  -f-  600 
Mean  temperature  of  gas  =  —  — ^ —    -  =  975°  F. 

Temperature  of  steam  and  water  in  boiler  =  355.8°  F. 

0  =  H(975  +  355.8)  =  665.4 
Hydraulic  radius  =  H  =  1. 
Area  of  tube  =  ^  =  0.0872. 


=  [3.327  +  0.839]  (619.2)  =  2570  B.t.u. 
Now  ri45  =  865.2  B.t.u. 

Hence  the  pounds  of  steam  per  square  foot  =  ogc~o  =  2.97  Ibs. 

This  result  is  an  average  value  per  square  foot  for  the  whole  boiler.  The 
value  of  the  shell  directly  over  the  fire  increases  the  average  to  more  than 
this  value.  If  the  temperature  entering  the  flue  had  been  higher  this  value 
would  be  greater. 

For  water  tube  boilers  it  is  suggested  that  the  area  be  considered  the  area 
between  tubes  and  that  the  hydraulic  radius  be  taken  as  the  distance  between 
tubes. 

Problem  2.  —  Thus  if  the  gases  enter  the  tubes  at  a  temperature  of  1900° 

M 
F.  and  leave  at  600°  F.  and  if  the  tubes  are  3  in.  apart  and  the  value  of  — 

r 

is  the  same  as  before,  the  following  results  would  hold  for  a  water  tube  boiler. 
Mean  temperature  gases  =  1250°  F. 

</>  =  803 
Hydraulic  radius  =  3  in.          ~w  =  0.65 

-  355.8) 


=  [4.015  +  0.613]  (894.2)    =  4120  B.t.u. 
Weight  of  steam  =  ov      =  4.76  Ibs. 


Problem  3.  —  Air  at  350°  F.  is  to  be  cooled  to  75°  F.  in  an  intercooler 
made  up  of  %-in.  tubes  with  the  air  passing  through  the  tubes  and  cooled 


HEAT  TRANSMISSION  111 

by  water  entering  at  60°  F.  and  leaving  at  90°  F.     Find  mean  At  if  (a)  K 

Kf 
is  independent  of  temperature  and  (6)  if  K  =  777712* 

In  this  problem  it  is  evident  that  a  counter-current  flow  must  be  em- 
ployed since  the  water  leaving  the  intercooler  is  higher  than  the  air  leaving. 

AZi  =  350  -  90  =  260°  F. 
At2  =    75  -  60  =     15°  F. 

XT  OAZ  ^    l  M<i  +  A*2\        1  /260+15\       137.5 

Now  Ati  -  A/2  -  245  >  io(— a" )  =Io(— 2— j  =  "IT 

Hence  the  exact  methods  must  be  used. 

Ah  -  AZ2       260  -  15  245 

(a)  Mean  At  =  -    — - —  =  -    — ^^  =  =  86.2 

*1  20      2.3X1.237 


15 

81.7 


=95-5 


Problem  4.  —  Find  the  heat  per  square  foot  for  temperature  conditions  of 
problem  3  if  the  water  has  a  velocity  of  5  ft.  per  second  over  the  tube  and 
the  air  has  a  velocity  of  20  ft.  per  second.  The  pressure  of  the  air  is  70.3  Ibs. 
gauge. 

ra  for  air  at  85  Ibs.  abs.  and  at  35°^"    5  or  212.5°  F. 

85  X  144 
=  53.35  X  672.5  =  °'341  lbs' 

(a)  Using  (30)  and  (32)  : 


K  =  2+  0022  X  °'341  X  20  =  20'2 
Now  Q  =  K  (mean  At)  =  20.2  (86.2)  =  1740  B.t.u. 

(6)  Using  (41): 

f  r  q 

Q  =  3600  j  0.0015  + 1  0.000506  -  0.00045  X  ^  +  0.00000165- 


(20X0.341) 


(212.5  -  75)  =  21.6(137.5)  =  2970  B.t.u. 


(c)  Using  (42'): 


X  wau  =  0.01287(1  +  127  X  10~5  X  43)  =  0.0135 
X  gas  =  0.01287(1  +  127  X  10'6  X  180.5)  =  0.01583 
„  _  0.0135    T20  X  0.24  X  0.3411  o.786 

10      TiT4  [      0701583  —  J 


Q  =  15.0  X  86.2  =  1293  B.t.u. 
(d)  Using  (43)  and  (44): 

K  -  126  X  0.341  X(18.25)o-«(5)M 
(95.5)  W 

Q  =  36.6  X  95.5  =  3490  B.t.u. 


112  HE  A  T  ENGINEERING 

The  results  do  not  agree  very  well  and  result  (d)  is  very  high  with  (6)  next, 
while  result  (c)  is  very  low.  The  reason  for  the  low  value  of  (a)  is  due  to  the 
fact  that  the  data  from  which  equation  (32)  was  deduced  holds  for  smaller 
differences  of  temperature.  Equation  (32)  will  in  general  give  results  on  the 
safe  side  as  the  temperature  differences  for  which  it  holds  are  small. 

Problem  5.  —  A  sterilizer  operates  on  a  counter-current  principle  with  the 
warm  water  entering  at  212°  F.  and  leaving  at  75°  F.  and  the  cool  water 
entering  at  70°  F.  and  leaving  at  202°  F.  The  liquid  moves  at  a  velocity  of 

2  ft.  per  second.     Find  the  mean  At  :  (a)  if  K  is  constant,  (6)  if  K  =  ~rf^ 
and  (c)  arithmetic  mean  A£  of  At's  at  the  two  ends. 

A«i  =  212  -  202  =  10°  F. 
A«,  =    75  -    70  =    5°  F. 


. 

(a)  Mean  A<  =        ~*    =  ^       =  7'24 


Problem  6. — Find  K  and  Q  per  square  foot  for  problem  5. 
Use  (29)  with  K"  =  6  for  water  and  at  150°  F.,  mw  =  61.2: 


1    , 

Qmwww  """ 

Qm 

•wIVw 

Q 

=  367  X  7.24  =  2654  B.t.u. 

.a,  (55) 

K 

60              171 

2 

1+-7TA/2 

Q  =  171  X  7.24  =  1240  B.t.u. 

Problem  7.  —  If  the  walls  are  He  in.  thick  in  the  sterilizer  what  is  the 
drop  in  temperature  in  the  copper  of  the  tube  to  transmit  the  heat  of  prob- 
lem 6? 

Using  equation  (3)  and  the  tabular  value  of  C  for  this: 

239.0(1  +  0.00003X150)  {fc  _  y 


16  X  12 

'•-'•-  fexISlhsi  -  ao27° 

Even  in  this  case  the  fall  of  temperature  occurs  mainly  in  the  films  of 
water. 

Problem  8.  —  The  temperature  of  the  condensing  water  is  70°  F.  at  inlet 
and  80°  F.  at  outlet.  The  temperature  of  the  steam  is  105°  F.  Find  mean 
A£  and  the  heat  removed  per  square  foot  of  cooling  surface  of  admiralty 


HEAT  TRANSMISSION  113 

metal  if  the  surface  is  clean  and  the  pressure  in  the  steam  space  is  1.2  Ibs. 
absolute,  and  the  velocity  of  the  water  is  4  ft.  per  second. 

A«i  =  105  -  70  =  35°  F. 
A£2  =  105  -  80  =  25°  F. 

|-H(35-  25)1*      f         1.25         1* 
Mean   A*  =    [35^26*]      =  Ll.560-l.49eJ 


PS       1.098 
*  =  pt  =  L200 

630  XI  X0.9142x0.98X\/4 

(29.9)  H  673  Rtu- 

Q  =  673  X  29.9  =  2010 

Problem  9.—  Steam  at  215°  F.  is  used  to  heat  water  at  60°  F.  to  180°  F. 
with  the  steam  inside  of  copper  pipes  H  in.  in  diameter  and  80  in.  long. 
The  velocity  for  the  water  is  2  ft.  per  second. 

Using  Orrok's  formula  : 

Mi  =  215  -  60    =  155°  F. 
A£2  =  215  -  180  =    35°  F. 


.,  fK(155-35)"  eo 

Mean   A'  =  =  l.56 


9*     f          15 
=  Ll.878- 


630  X  1  X  1  X  l.OOA/2 

K  = 


Q  =  513  X  82  =  42,100  B.t.u.  per  square  foot. 
Using  equation  (51)  of  Hagemann's, 


=  10  +  441  =  451 
Q  =  451  X155  +  35  =  42,800  B.t.u.  per  square  foot. 

Problem  10.  —  Suppose  the  pipe  in  problem  9  was  used  to  boil  water  at 
180°,  find  K  and  Q. 

Using  equation  (52)  of  Jelinek: 

K  _  _mo==  .  1965 

HsV*  X80 

Q  =  1965  X  (215  -  180)  =  68,800  B.t.u. 

Problem  11.  —  A  brine  cooler  has  brine  with  a  velocity  of  1  ft.  on  one  side  at 
20°  to  10°  F.  and  ammonia  on  the  other  at  0°  F.  Find  K  and  Q  per  square 
foot. 

In  this  problem  the  method  will  be  to  use  the  same  constants  as  those 
used  for  steam  to  water.  Using  Hagemann's  equation  (51): 


10  +  119  =  129 

129  X  15  =  1935  B.t.u.  per  hour  per  square  foot. 


114 


HEAT  ENGINEERING 


For  4  ft.  velocity  this  would  give  for  K 

K  =  10  +  238  =  248 

Problem  12. — An  ammonia  condenser  uses  steel  pipes  with  water  from 
60°  to  80°  F.  and  ammonia  at  90°  F.     The  water  velocity  is  4  ft.  per  second. 
Using  equation  (50) : 

K  =  130  X  VI  =  260 

Q  =  260  X  23"^ — 3  =  260  X  18.3  =  4750  B.t.u. 

Problem  13. — Find  the  number  of  sections  required  and  the  average  heat 
transmitted  per  square  foot  for  vento  heaters  to  operate  with  steam  at  220° 
F.  (5  Ibs.  gauge)  and  to  heat  air  from  60°  F.  to  114°  F.  when  delivered  across 
heater  at  1200  ft.  per  minute. 

From  Fig.  30:  Two  sections  will  heat  zero  air  to  60°  F.  and  five  sections 
will  heat  zero  air  to  115°  F.  at  1200  ft.  per  minute;  5  —  2=3  sections  are 
required. 

From  Fig.  31:  The  average  transmission  for  two  sections  with  zero  air 
is  2050  B.t.u.  while  with  five  sections  1600  B.t.u.  are  transmitted.  The 
amount  for  the  last  three  of  the  five  sections  will  be 

5  X  1600  -  2  X  2050       1Qnnn. 
Average  transmission  =  —  — ^ —  =  UUU  B.t.u. 

To  find  the  result  of  this  problem  by  formulae  in  place  of  using  the 
experimental  curves  the  following  is  given: 

60  -I-  115 
Mean  At  =  220  -  — ^ =  132.5 

1^  =  9.83  -; 

Q  =  9.83  X  132.5  =  1300  B.t.u. 


^Cement 
Plaster 


Air  Space 

FIG.  34. — Section  of  wall  for  cold  storage  room. 


Cork 
foaml 


Problem  14. — Find  the  value  of  K  for  the  wall  of  a  cold  storage  warehouse 
shown  in  Fig.  34. 

T  =  16.2  -  4.00  X  3  =  4.2 


a  for  outside  =  1.23  +  0.74  + 

=  2.28 

a  for  air  space  |    =  0<82  +  0<74  + 
a    for  inside  J 

=  1.79 


42  X  1.23+31  X0.74 
1000 

42  X  0.82 +31  X0.74 
1000 


X4.2 
X4.2 


HEAT  TRANSMISSION  115 

K 1 

i  2  1          J_       0.67      033      0.17         1 

~T~  n  /i «    i    1  >7(\  T  i  »7O    i    n/miniTin/ifti 


2.28^0.46^1.79   '  1.79    '    0.46   '  0.17   '  0.46   r  1.79 

1 

~  0.438  +  4.35  +  0.558  +  0.558  +  1.455  +  1-94  +  0.37  +  0.558 
=  0.0978 

It  will  be  noted  that  the  amount  of  air  space  does  not  aid  in  the  insulating 
value  of  the  wall  because  the  air  can  and  will  circulate.  The  value  of  this 
lies  in  the  resistance  at  the  two  surfaces  which  amounts  to  almost  as  much 
as  8  in.  of  brickwork.  The  various  terms  of  the  denominator  show  the 
values  of  the  various  elements  of  the  wall  as  heat  insulators.  Thus  4  in. 
of  cork  is  as  valuable  as  11  in.  of  brickwork  and  may  be  used  when  space 
is  of  value.  The  wall  above  will  transmit  2.35  B.t.u.  per  square  foot  per 
degree  difference  in  24  hours.  For  a  temperature  difference  of  60°  F.  this 
amounts  to  141  B.t.u.  per  day  per  square  foot. 

Topics 

Topic  1. — By  what  methods  is  heat  transmitted?  Explain  the  pecul- 
iarities of  each  method.  Give  the  Stefan-Boltzmann  Law.  For  what  is 
this  used?  How  is  it  applied? 

Topic  2. — What  is  the  law  of  conduction  as  stated  by  a  formula?  Is 
the  coefficient  of  conduction  a  constant?  What  are  the  dimensions  of  this 
coefficient?  What  fixes  the  amount  of  heat  carried  by  convection?  Why 
is  convection  of  value  in  engineering  problems?  Is  the  heat  in  these 
problems  carried  by  convection  or  conduction? 

Topic  3. — How  may  it  be  shown  that  the  films  of  substance  at  the  surfaces 
of  a  partition  or  plate  offer  the  greatest  resistance  to  flow?  What  is  the 
effect  of  velocity  on  film  resistance?  Explain  why  this  is  true.  What  was 
the  work  of  Dalby?  What  was  the  work  of  Reynolds? 

Topic  4.— Prove  that 

Q  =  K'mw(t  -  0) 
Is  this  the  equivalent  of  Reynold's  expression 

Q  =  (A  +  Bmw)(t  -  e) 

M 

Show  that  -  =  rnw 

F 

Topic  6. — What  is  meant  by  hydraulic  radius?  Determine  the  mean 
temperature  difference  along  a  surface  if  K  is  constant. 


L 


Topic  6. — What  is  meant  by  the  constant  Kf     Prove  that 

if  K  =  J^_ 

Topic  7.— Prove  that 

K  =  77 


" L  i L  : —  4. 

ci      r  Cn     "^  cm    • 


116  HEAT  ENGINEERING 

To  what  does  this  relation  reduce  for  ordinary  transmission  through 
thin  partitions? 

Is  c  a  constant?  What  is  the  form  to  which  Nicolson  reduces  this  value  of 
K  above? 

1  1  1 

Topic  8. — Given:         -~  =  ~ I- 

K        6mgwg 

reduce        K  = 


Topic  9.  —  What  are  the  formulae  of  Nicolson,  Jordan  and  Nusselt?  Are 
these  formulas  applicable  to  the  same  conditions?  On  what  does  the  heat 
per  square  foot  per  degree  per  hour  depend?  Give  details  and  reasons. 

Topic  10.  —  For  what  conditions  is  the  Rensselaer  formula  applicable? 
For  what  is  Orrok's  formula  used?  On  what  does  the  K  of  Orrok's  formula 
depend?  On  what  for  the  Rensselaer  formula? 

Topic  11.  —  Is  Orrok's  formula  applicable  to  ammonia  condensers, 
evaporators  and  feed-water  heaters?  Are  special  formulae  given  for  these 
forms  of  apparatus?  Give  the  formulae  used  for  finding  K  for  these. 
What  is  the  value  of  K  for  direct  steam  radiators?  For  indirect  steam 
radiators? 

Topic  12.  —  Explain  the  method  of  finding  the  heat  transmitted  per  square 
foot  of  surface  per  hour  in  Vento  heaters  and  in  pipe  coils  used  as  indirect 
heaters. 

Topic  13.  —  Explain  the  method  of  finding  the  value  of  K  for  a  wall  or 
partition. 

Topic  14.  —  Derive  the  expression  for  the  efficiency  of  heat  transmission 


when  K  is  independent  of  the  temperature.  Give  all  steps  and  discuss  the 
variation  of  the  efficiency  of  a  surface  with  velocity,  diameter  of  pipe, 
length,  temperature,  and  heat. 

Topic  15.  —  Starting  with  Reynolds'  value  of  K 


K  =  K'(A  +  B 
reduce  the  expression  for  the  efficiency  of  a  heating  surface. 


_ 

=   /   —  € 


17    =         —  €  M  hr.h 

PROBLEMS 

Problem  1.  —  A  piece  of  glass  is  held  at  750°  F.  and  a  shield  covers  two- 
thirds  of  this.  The  surface  of  the  glass  is  2  sq.  ft.  The  surface  of  the 
shield  is  20  sq.  ft.  The  shield  is  held  at  125°  F.  by  a  water-jacket.  How 
much  heat  is  removed  by  the  water-jacket? 

Problem  2.  —  A  4-in.  boiler  tube  has  gas  entering  at  one  end  at  1400°  F.  and 
leaving  at  550°  F.  with  steam  at  120  Ibs.  gauge  pressure.  The  coal  is  burned 


HEAT  TRANSMISSION  117 

on  a  grate  of  eight  times  the  area  through  the  tubes  at  a  rate  of  15  Ibs.  per 
hour  with  30  Ibs.  of  air  per  pound  of  coal.  Find  the  value  of  K  and  the 
number  of  B.t.u.  per  square  foot  of  surface.  How  many  square  feet  of 
surface  would  be  required  per  boiler  horse-power?  (One  boiler  horse- 
power equals  the  evaporation  of  34}4  Ibs.  of  water  at  212°  F.  per  hour  into 
dry  steam  at  212°  F.) 

Problem  3. — A  feed-water  heater  uses  steam  at  3  Ibs.  gauge  pressure  to 
heat  6000  Ibs.  of  water  per  hour  from  60°  to  200°  F.  The  water  is  passed 
through  at  a  velocity  of  2  ft.  per  second.  Find  mean  At  assuming  K  con- 
stant. Find  K  by  two  formulae.  Find  the  square  feet  of  surface  necessary. 
Of  what  material  will  the  tube  be  made?  Why? 

Problem  4. — An  economizer  is  used  to  heat  6000  Ibs.  of  water  per  hour 
from  60°  F.  to  200°  F.  by  using  hot  gas  at  500°  F.  in  3-in.  iron  flues  in  which 
the  temperature  is  reduced  to  350°  F.  The  velocity  of  the  gas  is  25  ft. 
per  second.  What  is  the  value  of  mean  At?  Find  K  by  Nusselt's  formula. 
Find  the  amount  of  surface  required.  Also  use  the  simple  formula 
suggested  by  Nicolson  in  which  L  is  employed. 

Problem  5. — In  a  condenser  the  water  enters  at  50°  F.  and  leaves  at  65° 
F.  with  a  pressure  in  the  condenser  of  0.6  Ibs.  absolute  and  a  temperature  of 
80°  F.  What  is  the  value  of  mean  At  for  this  case?  Find  K  for  copper 
tubes  if  dirty.  Find  the  square  feet  per  kilowatt  of  turbo  generator  if 
steam  consumption  is  14  Ibs.  with  x  of  exhaust  steam  equal  to  0.95. 

Problem  6. — Find  the  size  of  the  condenser  for  an  ammonia  plant  to 
remove  200,000  B.t.u.  per  hour  with  water  flowing  at  5  ft.  per  second  in  the 
double  pipe,  entering  at  65°  F.  and  leaving  at  80°  F.,  when  the  ammonia  is 
at  100°  F. 

Problem  7. — A  boiler  using  a  24-in.  flue  has  gas  entering  at  1600°  F. 
and  leaving  at  1000°  F.  The  steam  is  at  300°  F.  The  velocity  of  the  gas 
is  100  ft.  per  second.  Find  the  heat  transmitted  per  hour  per  square  foot 
of  flue. 

Problem  8. — An  intercooler  receives  air  at  250°  F.  and  cools  it  to  80° 
F.  The  water  enters  at  50°  F.  and  leaves  at  70°  F.  Find  mean  At  for 

•rrt 

parallel  and  counter-current  flow  using  K  =  constant  and  K  =  —?.     Find 

the  heat  transmitted  per  square  foot  per  hour. 

Problem  9. — Find  the  surface  required  in  an  interchanger  cooling  7000 
Ibs.  of  water  per  hour  from  220°  F.  to  80°  F.  by  water  entering  at  60°  F.  and 
leaving  at  200°  F.  Use  K  =  constant. 

Problem  10. — Find  the  surface  required  to  boil  500  Ibs.  of  solution  at 
200°  F.  by  steam  at  250°  F.  if  tubes  3  ft.  long  and  3  in.  in  diameter  are 
used  and  the  heat  of  vaporization  of  the  liquid  is  750  B.t.u. 

Problem  11. — Assume  air  at  50°  F.  and  move  it  with  a  velocity  of  1200 
ft.  per  minute  over  the  Vento  heaters  or  coils  to  heat  it  to  105°  F.  How 
many  sections  will  it  take  for  the  Vento  heaters  and  for  the  coils?  How 
many  square  feet  of  each  will  be  required  to  heat  200,000  cu.  ft.  of  air  per 
hour? 

Problem  12. — Find  the  K  for  a  wall  composed  of  16  in.  of  brickwork,  a 
2-in.  air  space  and  12  in.  of  brickwork  with  1  in.  of  cement  plaster. 


CHAPTER  IV 
AIR  COMPRESSORS 

Compressed  air  is  used  for  many  purposes  for  which  it  would 
be  difficult  to  employ  other  media.  For  operating  small  tools, 
rock  drills,  and  hoists  and  for  the  transmission  of  power  over  con- 
siderable distance  it  replaces  steam  with  which  the  loss  due  to 
condensation  is  very  excessive.  The  efficiency  of  transmission 
of  power  by  compressed  air  is  not  equal  to  that  of  electrical 
transmission  nor  is  it  as  flexible,  yet  in  certain  cases  for  some 
reasons  it  is  of  value.  For  cleaning  materials  with  a  sand  blast 


FIG.  35. — Section  of  radial  blade  fan 
blower. 


FIG.  36. — Rotary  blower,  section. 


and  for  use  with  the  cement  gun  compressed  air  is  necessary. 
To  force  air  into  a  furnace  of  a  boiler  or  for  a  blast  furnace  a 
compressor  of  some  form  is  required.  Compressed  air  may  be 
used  in  air-lift  pumps  and  in  direct  air  pumping.  In  supplying 
air  to  divers,  in  the  making  of  liquid  air  and  in  the  operation 
of  street  cars  compressed  air  is  required. 

To  compress  this  air  several  types  of  compressors  are  used. 
For  low  pressures  up  to  10  oz.  per  square  inch  above  or  below  the 
atmosphere,  centrifugal  or  fan  blowers,  Fig.  35,  are  used.  In 
these,  radial  or  curved  blades  are  driven  at  a  high  speed  causing  a 

118 


AIR  COMPRESSORS 


119 


flow  of  air.  These  are  used  for  ventilation  of  buildings;  for  supply 
of  air  to  boiler  furnaces  or  cupolas;  for  forges;  for  suction  in  an 
induced  draft  system  or  for  the  conveying  of  light  materials. 
In  forge  work  Sangster  allows  140  cu.  ft.  of  free  air  per  minute 
per  forge  at  2  oz.  pressure.  If  an  exhaust  fan  is  used  600  cu.  ft. 
of  air  per  minute  at  %  oz.  pressure  are  handled  per  forge.  The 
air  required  in  cupolas  is  40,000  cu.  ft.  per  ton  of  iron  melted. 
For  ventilation  2000  cu.  ft.  of  air  are  allowed  per  person  per  hour. 

For  pressures  of  from  8  oz.  to  7  Ibs.  per  square  inch,  rotary 
blowers  such  as  that  shown  in  Fig.  36  are  used.  These  are  run 
at  a  sufficient  speed  to  get  the  necessary  discharge  in  cubic  feet. 

For  pressures  up  to  35  Ibs.  turbo  compressors  of  the  form  shown 
in  Fig.  37  are  used  although  piston  compressors  are  used  for  this 
pressure  at  times.  The  air  at  this 
pressure  is  used  in  blast  furnaces 
and  converters  and  the  compressors 
of  the  piston  type  for  such  are 
known  as  blowing  engines  or  blow- 
ing tubs.  These  are  only  special 
forms  of  air  compressor. 

For  higher  pressures  than  35  Ibs. 
the  piston  compressor  is  in  general 
employed,  although  the  hydraulic 
compressor  or  the  Humphrey  explo- 
sion compressor  could  be  used.  For 
high  pressures  it  will  be  shown 
that  efficient  work  necessitates  the 
compression  in  one  cylinder  to  a 
pressure  much  below  that  desired  and  then  a  further  compres- 
sion of  this  air  to  a  higher  pressure.  It  may  be  that  a  third 
compression  is  used  before  the  desired  pressure  is  reached.  Each 
one  of  these  cylinders  is  known  as  a  stage.  The  compressor  men- 
tioned above  would  be  known  as  a  two-stage  compressor.  In 
general  one  stage  is  used  to  70  or  80  Ibs.  gauge  pressure  while 
from  that  to  500  Ibs.  two  stages  are  used,  and  three  stages  from 
500  to  1500  Ibs.  Above  this  four  stages  would  be  used.  Fig. 
38  shows  a  two-stage  compressor.  In  this  air  is  sucked  into  the 
center  of  the  piston  A  by  the  vacuum  produced  behind  the  piston 
when  the  piston  moves  to  the  left,  the  air  flowing  through  an  open- 
ing left  at  the  periphery  B,  as  shown  in  Fig.  39.  The  air  on  the 
left  of  the  piston  is  compressed  and  after  it  reaches  the  pressure 


37.-Section  of  turbo  blower. 


120 


HEAT  ENGINEERING 


existing  above  the  mushroom  valve  at  C  this  opens  and  allows 
the  air  to  exhaust  into  the  discharge  pipe  D  and  from  this  into 
the  intercooler  E  in  which  the  air  is  cooled  by  water  in  the  pipes  F. 
The  water  is  caused  to  circulate  back  and  forth  in  this  inter- 


FIG.  38. — Section  of  two-stage  Ingersoll-Rand  compressor. 

cooler.  The  air  just  compressed  in  cylinder  G  is  forced  through 
the  intercooler  into  the  cylinder  H  where  it  is  compressed  to  a 
higher  pressure.  In  this  cylinder  the  inlet  valves  I  are  at  the  top 
of  the  cylinder  while  the  discharge  valves  J  are  at  the  lower  part 
of  the  cylinder  head.  Both  sets  of  these  are  mushroom  valves. 


r*^;;SvV;^ 

,ly£vV viii/VA.'-.v ^'-'- :;£:.- .•A'^;-'4v. v •.'•'.-.  '^',- -'^ ^ :^' '<7-v'f'-'--'*:v-':^:-.'o:>.:»^ 

). — Enlarged  section  of  L.  P.  air  cylinder  of  Inger; 


FIG.  39. — Enlarged  section  of  L.  P.  air  cylinder  of  Ingersoll-Rand  compressor. 

The  air  is  finally  sent  to  the  ak  storage   tank   or   aftercooler 
through  the  pipe  K. 

Fig.  38,  which  is  the  Ingersoll-Rand  class  AA-2  compressor, 
illustrates  one   form  of   two-stage  air  compressor  in  which  the 


AIR  COMPRESSORS 


121 


driving  steam  cylinder  0  is  in  line  with  the  two  air  cylinders 
and  the  fly  wheels  are  driven  by  outside  connecting  rods.  The 
figure  illustrates  the  method  of  bringing  cool  air  to  the  compressor 
from  a  point  outside  of  the  engine  room  through  the  conduit  L 
and  at  M  and  N  are  the  water  jackets  to  remove  some  of  the  heat 
of  compression. 

Fig.  39  is  introduced  to  show  the  valves  of  the  low-pressure 
cylinder.     In  the  periphery  of  each  face  of  the  piston  are  a  series 

of  slots  distributed  around  the  piston 
through  which  air  can  pass.  Over  this 
is  fitted  a  ring  which  closes  this  open- 
ing and  acts  as  a  valve.  This  is  known 
as  the  hurricane  inlet  valve.  When 
the  piston  moves  to  the  left  at  the  be- 
ginning of  a  stroke  the  right-hand  ring 
is  moved  from  its  openings  and  air  can 
enter  behind  the  right  side,  the  com- 
pressed air  on  the  left  holding  closed 
the  valve  on  the  left  side.  At  the  end 
of  the  stroke  the  inertia  of  the  ring 
tends  to  close  the  right  one  as  the  pis- 
ton reverses  and  the  left  one  will  open 
as  soon  as  the  pressure  of  the  air  in 
the  left-hand  clearance  space  expands 
to  atmospheric  pressure.  The  dis- 
charge valves  are  ordinary  mushroom 


FIG.  40. — Vertical  after  cooler  and  intercooler  of  Igersoll-Rand  Co. 

valves    with    a   light  spring  to  close  them  and  a  tube  on  the 
back  to  act  as  a  guide. 

Fig.  40  shows  the  construction  of  an  intercooler,  through  which 
the  air  must  pass  from  one  stage  to  another  and  give  up  its  heat. 
By  the  arrangement  of  baffle-plates  and  partitions  the  air  and 
water  are  made  to  take  a  circuitous  path  so  as  to  be  more 
efficient  in  the  removal  of  heat.  The  moisture  which  separates 


122 


HEAT  ENGINEERING 


as  the  air  is  cooled  is  usually  caught  as  shown  in  the  figure  so  that 
it  will  not  pass  over  into  the  next  stage. 

In  many  cases  an  after  cooler,  Fig.  40,  is  used  after  the  last 
stage  to  remove  more  of  the  moisture  from  the  air  and  to  cool  it 
before  it  passes  into  the  transmission  line.  In  this  way  the  air 
is  of  smaller  volume  and  there  is  less  friction. 

Fig.  41  illustrates  the  arrangement  of  the  Taylor  hydraulic 
air  compressor.  In  this  a  system  for  the  flow  of  water  must 
exist.  Suppose  the  dam  A  gives  a  head  of  H  feet  and  this  causes 
water  to  flow  through  the  pipes  B,  C,  D,  E  to  the  tail  race  F. 
The  head  and  friction  will  cause  a  certain  flow  through  the  system 


FIG.  41. — Taylor  hydraulic  air  compressor. 

and  if  the  pipe  is  necked  at  C  the  velocity  may  become  so  high 
that  a  vacuum  is  formed  at  this  point  and  air  is  drawn  in  through 
openings  placed  here.  The  air  enters  from  G.  This  air  mixes 
with  the  water  and  when  the  velocity  is  decreased  in  H  the  air 
separates  out  and  rises  to  the  surface  of  the  water.  This  air  is 
under  pressure  due  to  the  head  h  on  the  chamber  G.  The  air  is 
taken  out  through  the  pipe  7. 

The  Humphrey  apparatus  is  described  in  Engineering, 
Vol.  LXXXVIII,  p.  737,  and  in  " Compressed  Air  Practice" 
by  Richards. 

In  all  of  these  forms  of  apparatus  the  volume  of  air  taken 
in  might  be  the  same  while  that  discharged  is  determined  by  its 
pressure  and  temperature.  To  give  some  idea  of  the  amount  of 


AIR  COMPRESSORS  123 

air  used  by  tools  or  machines  and  the  amount  handled  by  the 
compressor  it  is  customary  to  reduce  the  air  to  some  one  standard 
condition.  The  conditions  taken  are  14.7  Ibs.  absolute  pres- 
sure and  60°  F.  temperature.  This  air  is  known  as  "free  air." 
At  times  the  temperature  is  assumed  to  be  that  of  the  atmosphere 
at  the  time  of  use,  in  which  case  the  term  free  air  refers  to  air  at 
atmospheric  pressure  regardless  of  temperature. 

With  this  introduction  the  thermodynamics  of  compressed 
air  will  be  considered. 

WORK  OF  COMPRESSION 

The  amount  of  work  required  to  compress  V/  cu.  ft.  per  minute 
of  free  air  is  shown  by  the  diagram  of  Fig.  42  which  assumes  no 
clearance.  The  line  db  is  the  atmospheric  line  and  on  account 
of  the  friction  of  the  inlet  pipe  and  P 
valves,  the  initial  pressure  p\  is  be- 
low  atmospheric  pressure.  The  air 
is  sucked  in  on  the  suction  line  cl 
and  is  compressed  from  1  to  2  on 

the   line    pvn  =  const,   and   is  then  a  |  _  ^^^_    6 
driven  out  from  2  to  d.     The  pres- 
sure at  1  is  pi  and  the  volume  is  V\. 
The  atmospheric  pressure  is  p.  and 
the  free  air  V/  after  throttling  oc- 
cupies the  volume  Vi.     The  throttling  action  means    constant 
heat  content  which  for  a  perfect  gas  means  isothermal  action. 

Hence  paVf  =  p.V,  (1) 

Some  authors  say  that  the  air  changes  its  temperature  in 
entering  the  cylinder,  but  this  cannot  be  appreciable  as  air  is 
such  a  poor  conductor  of  heat  that  it  would  take  up  little  or  no 
heat  from  the  walls  and  the  throttling  of  itself  is  isothermal. 

Now  the  work  of  compression  is 

=  W 


i 


This  quantity  is  negative  since  the  work  is  done  on  the  gas. 
negative  answer  means  that  work  is  done  on  the  gas. 
Reducing: 


124 

Now 

Hence 


HEAT  ENGINEERING 


work  = 


n 


—  ( — )   n 

\Vi/ 


(3) 


If  desired  paVf  may   be  substituted   for   p\V  i  giving    work 
to  compress  V/  cu.  ft.  of  free  air  from  pi  to  p2  as 


(4) 


EFFECT  OF  CLEARANCE 

Now  if  there  is  clearance  the  indicator  card  takes  the  form 
shown  in   Fig.   43.     The   air  remaining  in   the   cylinder,   d2', 

at  the  end  of  discharge  expands 
from  2'  to  1'  on  the  return  stroke 
preventing  air  from  entering  un- 
til I/  is  reached.  The  amount 
of  air  then  taken  in  is  V"\  — 
V'i,  or  V"i  -  Vi  =  Vi  of  the 
previous  discussion. 

The  net  work  required  to 
drive  the  compressor  is  area  1" 
2"2'1'  or  area  l"2"dc  -  1'2'dc. 
This,  assuming  the  same  form 


FIG.  43.-Effect  of  clearance. 


of  expansion  and  compression  lines,  gives 
Net  work  with  clearance  = 


<» 


This  is  the  same  expression  as  (3)  for  the  compression  of  V\ 
cu.  ft.  of  air  from  pressure  p\  to  pz,  or  in  other  words  clearance 
has  no  effect  on  the  work  required  to  compress  a  definite  volume 
of  air.  The  effect  of  clearance  is  to  decrease  the  amount  of  air 
taken  in  for  a  given  displacement  or  to  increase  the  displacement 


AIR  COMPRESSORS  125 

for  a  given  amount  of  air  taken  in.  Thus  in  Fig.  42  the  air  taken 
in  is  Vi  and  the  displacement  D  =  Vi,  but  in  Fig.  43  the  air 
taken  in  is  Vi  but  the  displacement  D  =  V\  +  1'c  —  2'd,  a 
quantity  greater  than  V\.  In  other  words  the  displacement  has 
been  increased.  This  increase  of  displacement  means  a  larger 
cylinder  in  stroke  or  area  of  piston  and  hence  there  will  be  more 
friction  and  consequently  more  work  will  be  required  to  drive 
the  compressor,  but  the  work  indicated  on  the  air  for  the  com- 
pression of  a  certain  amount  is  the  same  with  or  without 
clearance. 

The  volume  2'd  represents  the  volume  of  the  clearance  space 
and  since  this  is  usually  expressed  as  a  percentage  of  the  dis- 
placement it  may  be  represented  as  ID,  where  I  is  the  percentage 
clearance  and  D  is  the  displacement. 


Now 

pi 


»© 


Hence  V  i  =  D  +  ID  -  ID          » 

(6) 


+  I  —  I  (—  j  n  i 


The  term  1  +  I  —  I  —  j  n  is  called  the  clearance  factor.     It 

may  be  represented  by  KI.  This  term  is  only  used  to  find  the 
displacement  if  Vi  is  given  or  Vi  if  D  is  given.  The  clearance 
ratio  I  is  known. 

EFFECT  OF  LEAKAGE 

If  there  is  leakage  around  the  piston,  piston  rod  and  through 
the  valves  the  volume  Vi  required  to  deliver  a  given  amount 
of  free  air  V/  must  be  increased  so  as  to  care  for  this  leakage. 
If  the  amount  of  free  air  delivered  is  expressed  as  a  percentage  of 
the  free  air  taken  in  or  as  a  ratio  to  the  free  air  taken  in,  the  per- 

centage is  called  the  leakage  factor  /.     Actual  V\=  -   ^-7— 

Although  the  leakage  is  effective  during  the  compression,  giv- 
ing less  work  as  the  compression  is  carried  higher,  it  may  be 
considered  that  this  loss  occurs  at  the  upper  pressure  only  and 


126  HEAT  ENGINEERING 

Ft         Vt 
consequently  increases  the  work  by  causing   y-  or  -4-  to  be 

substituted  for  Fi  or  F/.     Thus 

WOrk   =  —    — r-  pi  -^~     1    —   ( — )     n     I    = pa  ——     1    —   I  —  J     n 

n  —  1        /  L          Vpi/        J        n  —  1        j  L          \pi' 

(7) 
The  effect  of  leakage  is  to  increase  the  work. 

VOLUMETRIC  EFFICIENCY 

The  volumetric  efficiency  is  denned  as  the  ratio  of  the  free  air 
delivered  to  the  displacement  of  the  compressor.  If  the  actual 
free  air  is  used  this  is  the  actual  volumetric  efficiency  while  if 

the  indicated  free  air  is  used,  the  indi- 
cated volumetric  efficiency  is  obtained. 
Thus  if  ab  is  the  atmospheric  line  from 
actual    compressor,   indicated  volu- 

•};     AA     „    -,r       ~^    metric  efficiency  or  apparent  volumetric 
FIG.  44. — Card  from  com- 

dng  volumetric    Affi^PTW.v  =  *H  =  W. 

D       ab 


Now  '-    xy  =  .       - 

since  the  pressure  is  lowered  by  throttling  action  in  which  T 
is  constant.  The  temperature  of  the  cylinder  walls  may  change 
the  temperature  of  the  air  slightly  but  this  is  so  slight  that  the 
air  is  considered  to  be  at  inlet  temperature.  Substituting  for 
xy  its  value,  the  following  is  obtained 


Apparent  vol.  eff.  =  ^  =  ^  [l  +  I  -  I  (^J  *J  (8) 

actual  Vf      /  X  ind.  free  air 
True  vol.  eff.  =  — ^—  -  =  -       — ^ 

True  vol.  eff.  =  —  X  leakage  factor  X  clearance  factor.    (10) 

HORSE-POWER  AND  POWER  OF  MOTOR 

If  Fi  cu.  ft.  are  required  per  minute  the  expression  (7)  gives 
the  work  per  minute  in  foot-pounds.  Consequently  dividing 
the  expression  by  33,000  gives  the  horse-power  shown  by  the  air 


AIR  COMPRESSORS  127 

card.  This  result  must  be  divided  by  the  efficiency  of  the  air 
compressor,  about  85  per  cent,  to  95  per  cent,  depending  on  size, 
before  the  horse-power  to  apply  to  the  compressor  is  determined. 
To  find  the  horse-power  applied  to  the  motor,  be  it  a  steam  engine 
or  an  electric  motor,  the  above  result  must  again  be  divided  by 
the  efficiency  of  the  motor.  This  may  be  about  90  per  cent. 
Hence,  horse-power  applied  to  compressor  forFi  cu.  ft.  per  minute 


_J±  __  PiVi          fi       (P^\n-~ 
"n-  1  eff.  X/X  33000  L1    "  \p 

indicated  horse-power  of  engine  drive 

n      ___  _  1  _ 
n  —  1  eff.  compressor  X  eff.  of  engine 


TEMPERATURE  AT  THE  END  OF  COMPRESSION 

If  the  temperature  is  TI  at  the  beginning  of  compression  the 
temperature  at  the  end  of  compression  is 

t-i 


This  is  seen  from  the  following 

pzVf 
BT2\n 


COMPRESSION  CURVE 

The  compression  curve  desired  for  air  compression  depends 
on  the  way  in  which  the  air  is  to  be  used  after  compression.  Air 
is  such  a  poor  conductor  of  heat,  and  at  60  or  100  revolutions 
per  minute  the  action  of  the  compressor  is  so  rapid,  that  expan- 
sion in  an  engine  or  compression  in  the  compressor  practically 
takes  place  along  an  adiabatic 

pF1-4  =  const. 

If  air  is  compressed  along  an  adiabatic  1-2,  Fig.  45,  and  is 
expanded  along  an  adiabatic  in  the  engine  this  adiabatic  will  be 


128  HEAT  ENGINEERING 

2-1,  if  there  is  no  leakage  nor  cooling  between  the  compressor 
and  engine.  If,  however,  the  air  is  stored  in  a  tank  for  some  time 
before  using  in  the  engine  the  air  at  a  high  temperature  T2  is 
cooled  to  the  original  temperature  TV  This  causes  the  volume 
to  decrease  so  that  the  volume  occupied  by  the  air  in  the  cylinder 
of  the  engine  is  Vz  where  2'  lies  on  the  isothermal  12'.  The 
expansion  line  in  the  engine  is  now  2'!'  and  the  area  122'!' 
represents  the  loss  of  work  due  to  the  cooling  in  the  tank.  Al- 
though 1-2  is  the  best  line  for  compression  if  the  air  is  to  be  used 
before  it  can  cool  so  that  it  will  expand  in  the  engine  along  2-1 
giving  no  loss,  it  is  evident  that  1-2'  would  be  the  better  line  if 
the  air  is  to  be  stored  before  using,  since  the  temperature  along 
this  line  is  constant.  Hence  it  is  often  stated  that  isothermal 
compression  is  the  ideal  and  best  method  of  compression.  This 
is  true,  and  true  only,  if  storage  of  air 
is  to  be  employed  in  the  system,  for 
other  cases  adiabatic  compression  may 
.  const,  be  the  best  method.  If  isothermal  com- 
pression is  used  the  work  of  compression 
becomes  12'dc,  resulting  in  a  saving  in 

work  of  the  area  122'.     The  loss  when 
FIG.    45.  —  Saving  due  to 
cooling  on  compression,      the   expansion   in   the  air  engine  takes 

place  is  then  12'!'  instead  of  122T.     The 

expression  for  the  work  with  isothermal  compression  without 
clearance  is 


Work  =  plVi  loge       -  ptV't  + 
p% 

but  p^V'z  =  piVi 

:.  Work  =  P1V!  loge-  =  -  plVi  loge     -  (14) 


To  approach  this  isothermal  line  in  compression  a  water 
jacket  is  placed  around  the  cylinder  to  remove  heat,  or  water  or 
oil  is  sprayed  into  the  cylinder  to  reduce  the  heat.  These  methods 
are  not  very  effective  since  n  is  changed  only  from  n  =  1.4  to 
n  =  1.35.  This  saving  is  slight.  The  reason  for  this,  as  stated 
before,  is  the  fact  that  the  air  is  a  poor  conductor  and  also  it  is 
in  contact  with  the  cylinder  walls  a  very  short  time. 

HEAT  REMOVED  BY  JACKET 

The  heat  removed  by  the  jacket  is  made  up  of  two  parts,  that 
during  the  part  of  the  stroke  1-2  and  that  during  the  part  2-d. 


AIR  COMPRESSORS 


129 


const. 


An  expression  may  be  written  for  the  first  part  but  no  expression 
can  be  written  for  the  part  during  the  time  that  the  piston 
moves  from  2  to  d.  The  temperature  difference  between  the 
air  and  the  jacket  water  is  greatest  dur- 
ing this  time,  but  the  cooling  surface 
in  contact  with  the  air  is  decreasing 
and  this  effect  would  tend  to  decrease  the 
cooling  effect  while  the  greater  tempera- 
ture difference  would  increase  the  effect. 
If  the  effect  is  assumed  to  be  the  same 
as  that  during  the  portion  of  the  stroke 
1-2,  the  total  effect  may  be  found  by  multiplying  the  effect 
Fi  1 


FIG.  46. — Card  from  com- 
pressor with  jacket. 


during  1—2  by 


~r 

—   V 


or 


Now  heat  removed  on  line  1-2  = 


k-  1 


1-n 


From  the  above  and  by  considering  the  leakage  factor  /,  the 
following  is  obtained: 
Heat  removed  by  jacket 

iy\      J/*  ~~|  /r\     ~\T      I*  frf\ 


i r j 

/P]\il_a  - 


(15) 


SAVING  DUE  TO  JACKET 

The  work   done  by  the   compressor   when  the   exponent  is 
changed  from  k  of  the  adiabatic  to  n  is 


Work= 


n  -  1      J     L  \p! 

The  work  for  adiabatic  compression  is 

Work  = 


-1     / 
The  saving  due  to  the  jacket  is  the  difference  of  these  or 


130 


HEAT  ENGINEERING 


WATER  REQUIRED  FOR  JACKET 

If  the  heat  removed  by  the  jacket  per  minute  is  found,  the 
water  required  is  determined  by  assuming  the  possible  range  of 
temperature  and  then  finding  the  water  by 


(17) 


~  _  heat   per  minute  from  jacket  in  B.t.u. 

Cr  =  -f  -f 

q  o  —  q  i 

G  —  weight  of  water  per  minute. 
q'0  =  heat  of  liquid  at  outlet. 
q'i  =  heat  of  liquid  at  inlet. 

MULTISTAGING   AND   INTERMEDIATE   PRESSURES 

Although  jacketing  is  used  the  slight  change  in  the  exponent 
does  not  give  a  great  saving  in  work  and  moreover  for  high  pres- 
sures Tr2  becomes  so  great  even  with  a  jacket 


that  the  lubricating  oil  is  apt  to  ignite  and  cause  an  explosion. 
To  prevent  this  and  to  save  work  the  air  is  compressed  to  a 


pv  n  m  Const. 


FIG.  47. — Two-stage  compression. 

pressure  less  than  that  finally  desired  and  discharged  from  the 
cylinder  into  a  chamber  containing  a  number  of  tubes  carrying 
cold  water.  This  chamber  is  called  an  intercooler  and  the  tubes 
are  made  of  such  an  area  and  arranged  in  such  a  manner  that 
the  air  is  brought  to  its  original  temperature  at  1  before  it  is 


AIR  COMPRESSORS  131 

sent  to  a  second  cylinder  of  proper  volume  in  which  it  is  com- 
pressed to  its  final  pressure.  This  method  of  compressing  in 
two  cylinders  of  different  sizes  is  known  as  two-stage  compression. 
The  action  is  shown  in  Fig.  47.  Vi  cu.  ft.  of  air  are  drawn  into 
the  low-pressure  cylinder  and  compressed  to  a  pressure  p'z.  The 

(7/2\  n  —  1 
— )    »    is  then  discharged  from  this 

cylinder  and  through  the  intercooler  until  its  temperature  is 
reduced  to  TV  The  air  is  then  drawn  into  the  second  cylinder 
which  must  be  of  such  a  volume  as  to  take  the  air  which  occupies 
the  volume  V'\  found  on  an  isothermal  through  1  at  a  pressure 
pV  The  air  is  then  compressed  to  a  pressure  p%.  The  work  in 
the  two  cylinders  is  given  by 


^          r      ip'z\  n~1~i 

Work  in  low-pressure  cylinder  =  —  ^-r  piVA  1  —  (  —  )    n 

n  r        /  #2  \  re~i~| 

Work  in  high-pressure  cylinder  =      _  .,   pYF'il  1  —  Mr*]    w 


Now  plVl 

Hence 


(18) 


The  only  variable  in  this  expression  is  p'2.  Hence  to  find  the 
condition  for  minimum  work,  the  first  derivative  with  respect  to 
p'z  must  be  equated  to  zero. 


™  r      'W 1  /T)'  \ 

(total    work)   =  -  ~rPiFi ( — "I 

n  —  1^  n    \vi/ 


p  pi 


Pl'  Pi  \P2  P2 


=  1 

PlP2     . 

P'2    =    VjhP*  (19) 

That  is,  the  work  is  a  minimum  if  pr2  is  a  mean  proportional 
between  pi  and  p2. 


132  HEAT  ENGINEERING 

Substituting  this  value  of  p'2  in  (13)  the  following  results: 


If  there  are  three  stages,  proceeding  in  the  same  manner,  the 
equation  becomes 

Total  work  = 


Pl7  3 

—  1  '  \pi/ 

in  which  there  are  two  variables,  p'2  and  p"2.     The  conditions 

for  a  maximum  or  minimum  are  —  -    (total    work)p//2    =    0 
and  —  «r-  (total  work)  p>2  =  0.     These   give  p'2  =  Vpip"2    and 


p'j  = 


=    P12P2 


(21) 
(22) 


<  Pi2p2  — 

For  m  stages  there  will  be  m  —  1  intermediate  variable  pres- 
sures in  the  expression  for  total  work  and  there  will  be  m  —  1 
partial  derivatives  to  equate  to  zero  giving,  in  the  same  manner 
as  above, 

p'2  =  V^S     or  ^=(^}™  (23) 


P's 


Pi 


(24) 


P2 

It  is  to  be  noted  that  the  ratio  of  pressures  on  each  stage  is  the 
same  and  each  ratio  of  pressures  on  the  various  stages  is  equal  to 

( —  )  m.     If  these  are  substituted  in  the  expression  for  total  work 

the  equation  becomes 

Total  work  for  m-stage  compression  = 

mn       „  f-        /z>2\—  1        /25\ 


AIR  COMPRESSORS  133 

This  is  the  general  expression  for  the  work  of  an  m  stage  com- 
pressor on  the  compression  line  pvn  =  const.,  for  any  value  of  n 
except  1. 

INTERMEDIATE  TEMPERATURES 
Since  the  ratios  of  pressures  on  each  stage  are  equal  to  ( —  j  m 

it  follows  that  the  temperatures  at  the  end  of  compression,  T2,  are 
all  the  same  and  equal  to 


For  a  single-stage  compression  between  pi  and  p2 
T,  -  Ti  (&  -• 


HEAT   REMOVED    BY    INTERCOOLER 

In  the  intercooler  the  temperature  is  reduced  at  constant 
pressure  from  T2  to  TI  and  the  heat  is  given  by 

Heat  from  intercooler  in  foot-pounds  =  JMcp(T2  —  TI) 


T  , 

-    «/  r>rp     Cp(l  2   —    1  i 


If  leakage  is  considered  this  becomes 


AMOUNT  OF  WATER  FOR  INTERCOOLER 

In  the  above  section  the  heat  removed  by  the  intercooler  has 
been  determined.  The  amount  of  water  per  minute  required  for 
the  removal  of  this  heat  is  found  by  assuming  the  temperature 
allowable  at  inlet  and  the  temperature  at  outlet  desirable  and 
then  computing  by  the  formula, 

~  _  heat  per  minute  from  intercooler  in  B.t.u. 

q'o  ~  q'i 

G  =  weight  of  water  per  minute. 
q'o  =  heat  of  liquid  at  outlet. 
q'i  =  heat  of  liquid  at  inlet. 


134  HEAT  ENGINEERING 

The  area  of  the  surface  of  intercooler  is  found  by  methods  of 
Chapter  III. 

WATER  REMOVED  IN  INTERCOOLER 

If  air  at  pressure  pa  has  a  relative  humidity  pa  and  the  weight 
of  1  cu.  ft.  of  moisture  to  saturate  the  air  at  the  temperature  T\ 
is  m0,  the  total  amount  of  moisture  entering  per  minute  is 

Mw  =  pamaVf 

When  this  is  sent  to  the  intercooler  and  cooled  to  temperature 
Tij  after  it  is  compressed  to  pressure  p'2  and  volume  F'2  or  ^-f  F/, 
the  amount  of  moisture  held  in  the  air,  if  saturated,  will  be 

M'w  =  ra2  -/-  Vf. 
P2 

This  is  less  than  Mw  in  most  cases,  so  that  the  amount  of 
moisture  precipitated  is 

Mw  -  M'w 

If  M'w  is  greater  than  Mw  the  ratio  vW^  will  give  the  relative 

jJ/J.     iff 

humidity  of  the  air  leaving  the  intercooler.  This  same  method 
can  be  used  to  compute  the  moisture  removed  by  the  after  cooler. 

EFFECT  OF  LEAKAGE  IN  MULTISTAGE  COMPRESSIONS 

The  leakage  in  a  multistage  compressor  is  a  variable  quantity, 
the  leakage  making  the  amount  of  air  handled  by  the  various 
stages  different.  Thus  if  3  per  cent,  is  the  leakage  in  a  single 
stage  the  leakage  in  a  three-stage  compressor  might  be 

1  -  0.97  X  0.97  X  0.97  =  0.088  =  0.09  approximately. 

That  is,  the  amount  of  air  to  be  handled  by  the  lowest  stage  would 

y  '  YI 

be  /TqT,  that  by  the  second  stage  T^TT  and  that  by  the  third  would 

Fi 

be  TTq^.     Of  course  these  differing  amounts  of  air  would  change 

the  theoretical  discussion  above  but  to  a  slight  degree.  Because 
of  the  similarity  of  relations  for  various  stages  the  same  pressure 
ratios  will  be  used  although  the  works  on  the  various  stages  will 
not  be  the  same.  Hence  in  solving  various  problems  the  ex- 


AIR  COMPRESSORS  135 

pression  for  total  work  will  be  used  as  derived  before  substituting 
•TT  for  Vi,  and  using  as  /  the  mean  value  of  the  various  fs  for  the 

m  stages.'    In  computing  the  displacement  of  the  various  stages 
the  correct  value  of  /  for  each  stage  will  be  used. 


DISPLACEMENT  OF  CYLINDERS 

The  displacement  of  each  cylinder  can  be  computed  after  the 
leakage  factor  and  clearance  factor  are  known  for  the  cylinder. 

The  leakage  factor  has  been  discussed  in  the  previous  section, 
and  if  3  per  cent,  per  cylinder  is  assumed  the  various  leakage 
factors  up  to  four  stages  may  be  taken  as  88  per  cent.,  91  per 
cent.,  94  per  cent,  and  97  per  cent.  The  clearance  factor  is 
given  by 


or  =  1  +l-l~  (30) 

\PI/ 

If  the  clearance  is  the  same  on  each  stage  the  clearance  factor 
will  be  the  same  for  each  stage.  If,  as  is  often  the  case,  the  value 
of  I  is  small  for  the  low-pressure  cylinder  and  gradually  increases, 
the  value  of  Kt  will  gradually  become  larger  for  the  high-pressure 

cylinders.     Equations  (6)  and  (30)  show  that  as  —  or  —  becomes 

greater  Kt  becomes  less  and  so  the  effect  of  pressure  on  this  factor 
is  quite  noticeable.  With  very  high  pressures  on  any  stage  the 
volume  of  expanded  air  at  the  end  of  the  expansion  part  of  the 
stroke  is  so  great  that  only  a  small  quantity  of  air  will  be  drawn 
in.  This  is  another  reason  for  multistaging. 

Having  KI  and  /  for  any  stage  the  displacement  is  found  by 


n    - 

~ 


In  other  words,  the  free  air  to  be  handled  multiplied  by  the  ratio 
of  the  pressure  of  the  atmosphere  to  the  initial  pressure  on  the 
stage  considered  gives  the  volume  of  the  free  air  when  at  the 
pressure  of  this  stage  and  this  divided  by  the  volumetric  efficiency 
for  this  stage,  fxKix,  gives  the  displacement  per  minute  if  V/  is  the 
free  air  per  minute. 


136  HEAT  ENGINEERING 

If  now  the  number  of  revolutions  per  minute  is  known  and  if 
the  piston  speed  per  minute  allowable  is  assumed  the  length  of 
stroke  is  known  and  then  the  area  can  be  found  after  it  is  decided 
whether  or  not  the  compressor  is  double  acting. 

Piston  speed  =  2LN  =  200  to  700  ft.  per  minute.  (32) 

D  =  (Fh+  FC)LN.  (33) 

Fh  =  area  of  head  end  of  air  piston  in  square  feet. 
Fc  =  area  of  crank  end  of  air  piston  in  square  feet. 
L  =  length  of  stroke  in  feet. 
N  =  number  of  revolutions  per  minute. 

SIZE  OF  INLET  AND  OUTLET  PIPES  AND  VALVES 

The  valve  and  pipe  areas  are  such  that  the  velocity  of  air  is 
from  3000  to  6000  ft.  per  minute.  Although  these  are  high  the 
loss  in  pressure  is  not  excessive.  The  suction  valve  is  open 
during  a  longer  time  than  the  discharge  valve  and  for  that  reason 
it  seems  to  be  necessary  to  use  larger  areas  on  the  discharge 
valves.  On  the  other  hand,  the  effect  of  the  drop  is  more  notice- 
able at  the  lower  suction  pressure  and  therefore  the  suction  valve 
must  have  a  large  area.  The  valves  are  of  about  the  same  area. 
This  area  of  each  set  may  amount  to  8  per  cent,  of  the  piston 
area  for  piston  speeds  of  300  ft.  per  minute  while  for  speeds  of 
700  ft.  per  minute  12  per  cent,  might  be  used. 

The  pipes  connecting  cylinders  or  carrying  air  to  or  from  the 
compressor  should  be  designed  from  the  allowable  velocity  if 
short,  while  for  long  pipes  the  drop  in  pressure,  to  be  considered 
later,  should  determine  the  size. 

PERCENTAGE  SAVING  DUE  TO  MULTISTAGING  OVER  A  SINGLE 

STAGE 


n-l- 


Work  on  single  stage  =      _  ..  —^    1  —  f  —  j    » 
Work   on  multistage  = 


.'.  per  cent,  saving  =  100  -  100  ^  ~ P^  (34) 

Jm    I-,  /?>2\-^-| 


AIR  COMPRESSORS 


137 


This  saving  is  equal  to  the  area  Ii22'2i,  Fig.  48,  and  of  course 
it  is  the  difference  between  the  work  of  single-stage  compres- 
sion and  two-stage  compression.  The  saving  is  equal  to 


n 


n-1     f 


$- 


IT 

f 


(35) 


UNAVOIDABLE   LOSS    OF   COMPRESSION    ON   TWO    STAGES 

The  area  12ilil',  Fig.  48,  represents  an  area  which  cannot  be 
regained  by  a  single-stage  engine  and  may  therefore  be  called 
the  unavoidable  loss.  It  is  equal  to  12ied  —  1'lied. 


FIG.  48. — Saving  due  to  multistaging. 


This  is  divided  by  /  to  give  the  total  unavoidable  loss  with 
leakage.     For  three-stage  compression  the  loss  is  really  the  work 


138 


HEAT  ENGINEERING 


on  two  stages  minus  the  work  returned  from  one  stage  between 
pressures  p"2  and  pi. 


Loss  . 


pi 


pi 


This  is  shown  in  Fig.  49. 


FIG.  49. — Losses  and  gains  for  three-stage  compression. 
WORK  ON  AIR  ENGINE 

Air  engines  are  usually  of  one  stage  and  as  the  air  expands  its 
temperature  falls  so  that  at  the  end  of  expansion  the  moisture  in 
the  air  freezes  and  in  many  cases  clogs  up  the  exhaust.  To 
prevent  this  heat  may  be  applied  to  the  exhaust  pipe,  multistag- 
ing  may  be  used  as  shown  in  Fig.  51  or  the  air  may  be  initially 
heated  as  shown  in  Fig.  50  by  the  dotted  lines.  This  latter 
method  produces  such  an  increase  in  the  work  done  that  it  will 
be  carefully  investigated  later. 

The  pressures  between  which  the  air  engine  operates  are  p'z 
and  p'i.  These  are  different  from  p2  and  pi  because  there  is  a 
drop  in  pressure  due  to  friction  in  the  pipe  line  carrying  air  to 
the  engine  and  in  the  valves  entering  the  engine,  thus  changing 
Pz  to  p'z.  At  exhaust  the  back  pressure  must  be  above  the  atmos- 
pheric pressure. 

If  there  is  complete  expansion  in  the  engine  and  if  the  exhaust 
valve  is  so  timed  that  the  compression  is  complete,  there  is  no 
effect  of  clearance  on  the  work.  Complete  expansion  or  compres- 
sion means  the  carrying  out  of  these  actions  until  the  final  pres- 
sure is  reached  as  shown  in  Fig.  52.  The  effect  of  clearance  on 
the  displacement  is  the  same  as  that  in  the  compressor  and  the 


AIR  COMPRESSORS 


139 


same  formula  is  used.  In  all  theoretical  discussions  complete 
expansion  is  assumed  and,  for  the  present,  the  case  of  work 
or  quantity  of  air  for  an  engine  with  incomplete  expansion  and 
compression  will  not  be  discussed. 

The   expression   for   work   of  the   engine   without   clearance 
becomes,  from  Fig.  50, 


FIG.  50.  —  Single-stage   engine  card       FIG.  51.  —  Two-stage  engine 
with  expansion  line  after  preheating.  card. 


—  ~  --  p  —  -  —  p'\Vr\ 

1   —  K 


Work  of  single-stage  engine  = 


In  this  case  p'zV'z  have  been  factored  out  because  it  is  these 
two  terms  which  are  known  in  the  case 
of  the  engine  since  the  air  is  here  at  the 
original  temperature  TI;  while  at  the 
lower  pressure  the  temperature  is  low 
and  would  have  to  be  computed  by 


•  - 


jfc-1 
k 


FIG.  52. — Card  with  com- 
plete expansion  and  com- 

,     -  ,  T7/  T  T  i      f         i  pression  in  engine, 

before  piVi  could  be  found. 

It  is  to  be  observed  that  (^r)~fc~  is  less  than  unity  so  that  the 

\P  2/ 

expression  for   work  is   positive,  and   moreover  p'zV*  =  piV\ 
or  paVf  so  that  either  of  these  may  be  substituted,  giving 

k          r       /P'I\  k-  i-i 

Work  on  single-stage  engine  =  -7 r  paVf\  1  —  (—r)   *        (38) 

K  ~~   L  L  \p  2'  -1 

For  a  multistage  engine  the  work  becomes 

(39) 


140  HEAT  ENGINEERING 

If  leakage  is  considered  this  reduces  the  work,  giving 


The  intermediate  pressures  are  found  as  before 

LOSS  DUE  TO  COOLING  AFTER  COMPRESSION 

The  air  is  compressed  to  a  temperature  T2,  for  single  or  multiple 
staging,  and  on  storing  in  the  tank  or  pipe  line  its  temperature 
is  reduced  to  T\.  The  work  which  could  have  been  done  in  one 
stage  at  the  temperature  Tz  would  be  given  by 


while  after  cooling  to  TI  the  work  to  be  obtained  is 

Wo*  -j^ 

Hence  the  loss  is 

Loss  of  work  due  to  cooling 


In  this  the  first  bracket  is  due  to  the  compressor,  while  the 
second  refers  to  the  engine. 

UNAVOIDABLE  LOSS  DUE  TO  EXPANSION  LINE 

Another  unavoidable  loss  is  due  to  the  fact  that  the  expansion 
line  in  the  engine  is  of  the  form  pVh  =  const,  while  that  used  on 
the  compressor  has  been  of  the  form  pVn  =  const.  This  means 
that  with  a  single  expansion  engine  the  area  shown  in  Fig.  53 
by  abc  is  unavoidably  lost.  This  computation  is  made  before 
cooling  the  air  to  the  temperature  TI  and  while  it  is  at  the  point 
This  loss  is  given  by 


AIR  COMPRESSORS 


141 


n-l 


Since 


n  f    tpi\~r\ 

^^Il1     W  n  J 


(42) 


Const. 


FIG.  53. — Unavoidable  loss  from  difference  in  values  of  n  in  compressor  and 

engine. 

LOSS  OF  PRESSURE  IN  PIPE  LINE 

As  the  air  flows  through  a  pipe  the  pressure  is  decreased  due 
to  friction.  There  is  only  a  slight  change  in  velocity  and  hence 
this  action  is  that  of  throttling  in  which  the  heat  content  and 
consequently  the  temperature  of  the  perfect  gas,  air,  remains 
constant.  Now  it  is  found  that  the  drop  in  head  when  a  fluid 
flows  along  a  pipe  line  varies  with  the  length  of  the  pipe,  with  the 
square  of  the  velocity  and  inversely  with  the  diameter.  This  is 
usually  expressed  in  feet  of  head  of  substance  flowing.  Since  the 
velocity  w  of  the  air  is  only  constant  over  a  differential  length  of 
pipe  due  to  the  increase  of  volume  as  the  air  expands,  the 
differential  drop  in  pressure  is  all  that  can  be  expressed. 

—  Ctrl   =  ~y  p:      ul 

d  2g 
In  this :  6  is  a  constant  and  equal  to 

,   0.0274    ,    0.00145       0.0121 

0.0124  +  -        -  H ^ —  +  —, —  =  0.02  approximately.  (43) 

w  a  CLW 

w  =  velocity  in  feet  per  second. 
d  =  diameter  of  pipe  in  feet. 

1  Given  by  Green  Economizer  Co. 


142  HEAT  ENGINEERING 

g  =  32.2.  ft.  per  sec.  per  sec. 
dl  =  differential  length  in  feet. 

dh  =  differential  drop  in  head  in  feet  of  air.     (dh  is  negative 
because  it  decreases  as  dl  increases.) 

jy 

Now  -g7p  dh  =  dp 

T) 

^nF    =  weight  of  1  cu.  ft.  of  air 


MET  ^  1 
and  w  =    —  -  X  j 

F  =  area  of  pipe  in  square  feet. 
where  ,,  .  ,  ,  */  .  .  . 

M  —  weight  of  air  per  second  in  pounds. 

p    b  M2B2T*  dl 
Hence  -  dp  =  — 


r       b  M*BT 

-  J  Pdp-tj* 

(T  is  constant) 

Pi2  -  p22  _  b  M2  BT        bM2BT 

2          =d  F*    2gL~       **         =          d»  (44) 

2d 

(45) 
Mean  b"  =  28 

In  the  equations  (44)  and  (45),  pi  =  original  pressure  in 
pounds  per  square  foot,  p2  =  final  pressure  in  pounds  per  square 
foot,  L  =  length  in  feet  in  which  4drop  occurs.  The  formula  may 
be  used  to  find  p2,  M,  or  F,  depending  on  what  quantities  are 
given.  The  formula 

.  „  M2L 

'.       Pi2  -  2>22  =  b"  -- 


can  be  changed  to  refer  to  the  volume  of  free  air  at  60°  F.  instead 
of  weight  by  dividing  by  the  square  of 


53.35  X  520 
14.7  X  144    glvmg 


(46) 


AIR  COMPRESSORS  143 

Richards  suggests  &'"  =  Mooo>  if 

pi  and  p2  =  pressure  in  pounds  per  square  inch. 
Vf  =  cubic  feet  of  free  air  per  minute. 
L  =  length  of  pipe  in  feet. 
D  =  diameter  of  pipe  in  inches. 

•LtH      _ 

~  2000  * 
The  expression  (44)  may  be  simplified  as  follows: 


Pi  +  P* 

0       =  p  =  mean  pressure 

z 

BT 

—  =  mean  volume  of  1  Ib. 
P 


,~ 


or  Head  drop  in  feet  of  air  =     ,~  m  (47) 


where  wm  is  the  mean  velocity  since 

MET  V 


Equation  (47)  is  the  formula  used  in  hydraulics  where  the 
velocity  is  uniform. 

Equations  (44),  (45)  or  (46)  may  be  used  to  find  the  length  of 
pipe  for  a  given  drop  and  quantity  by  weight  or  volume  for  a 
given  pipe,  or  D  if  the  allowable  drop  in  a  certain  length  at  a 
given  discharge  is  known,  or  lastly  the  drop  for  a  given  discharge 
through  a  certain  pipe  of  given  length.  The  value  of  b  is  obtained 
by  successive  approximations  in  the  formula  (43)  if  not  known 
from  the  given  conditions. 

LOSS    DUE    TO    LEAKAGE    FROM    TRANSMISSION    LINE 

The  leakage  from  the  transmission  line  may  amount  to  a 
large  percentage  of  the  air  delivered  if  the  line  is  not  tight  through- 
out its  entire  length.  The  leakage  through  a  number  of  small 


144  HEAT  ENGINEERING 

holes  is  larger  than  one  would  expect.     The  amount  of  leakage 
is  found  by  Fliegner's  formulae,  depending  on  the  pressure. 

M=0.53  -^L  for  p2<  0.5  Pl 


M  =  1.060  F        alr~    2  for  p*  >  0.5  Pl 
After  M  is  found  this  may  be  reduced  to  free  air  by 


The  loss  is  proportional  to  the  quantity  of  air. 

Vfi 
Loss  as  per  cent,  of  what  should  be  obtained  =  100  -y-  • 

LOSS  DUE  TO  THROTTLING 

The  friction  action  of  the  pipe  line  is  the  equivalent  of  throttling 
action  and  moreover  in  many  cases  air  is  stored  in  tanks  under 
very  high  pressure  to  be  used  in  engines  or  air  motors  at  a  re- 
duced pressure  after  passing  through  a  reducing  pressure  valve. 
The  action  of  this  reducing  pressure  valve  is  throttling  action 
and  hence  the  temperature  of  the  air  remains  constant  during 
this  action;  p'zV'z  is  therefore  the  same  as  piVi  or  paV/.  The 
pressure  has  been  reduced.  The  available  energy  should  have 
been 

IP'--*- 


but  by  throttling  to  p'%  this  is  reduced  to 


or 

Loss  due  to  throttling  =  W  '  -  W"  = 

1  <«> 


This  reduction  has  been  due  to  the  reduction  of  the  upper 
pressure  to  a  point  near  the  lower  pressure. 


AIR  COMPRESSORS 


145 


GAIN  FROM  PREHEATING 

If  air  at  a  pressure  p'z  and  temperature  7\  is  heated  to  the 
temperature  T"'2  so  that  its  volume  is  changed  from  V2  to  ~F"2, 
the  heat  added  will  be 

AJQ  =  JMcp(T"z  -  T,) 
p'2(7"2  _  F'2) 


(49) 


FIG.  54.  —  Gain  from  preheating  air. 
The  increased  work  will  be  the  area  2'2"1"1',  or 


(50) 


The  efficiency  of  this  heat  used  in  preheating  is  therefore 


„ 


k  -  1 


"2  -  7'2) 


-(&)*?    (51) 


1      2  — 


If  now  this  efficiency  of  preheating,  1  —  f^r)    *  '     is   higher 

\p  2' 

than  the  overall  efficiency  of  the  system,  i.e.,  the  ratio  of  the  work 
of  the  air  motor  to  the  work  required  for  the  compressor,  it  will  pay 
thermodynamically  to  preheat.  Of  course  the  necessity  for  a 

warm  exhaust  or  the  possibility  of  increasing  the  output  of  a 

10 


146  HEAT  ENGINEERING 

given  quantity  of  air  may  make  it  advisable  to  use  preheating, 
although  the  efficiency  of  this  is  not  as  great  as  the  efficiency  of 
the  system. 

POWER  AND   DISPLACEMENT  OF  AIR  ENGINE   OR  MOTOR 

In  the  case  of  complete  expansion,  the  expression  for  work  has 
been  given. 

Work  for  single  stage  =  ~j  pafVf  [  1  -  (^)  HTJ          (38) 
Work  for  single  stage  after  preheating  = 

w^wk-ffi-?]    (52) 

Work  for  m-stage  expansion  = 


Since  in  these  the  volumes  represent  quantity  per  minute,  the 
indicated  horse-power  may  be  found  by  dividing  by  33,000  and 
this  quantity  multiplied  by  the  mechanical  efficiency  of  the  motor 
or  engine  will  give  the  delivered  horse-power.  This  efficiency 
may  be  taken  at  75.  to  90  per  cent. 

If  on  the  other  hand  the  power  required  is  known  as  well  as 
the  pressure  available,  the  indicated  power  can  be  found  from  the 
assumed  efficiency  and  then  the  work  per  minute.  From  the 
equations  above,  the  volume  of  free  air  necessary  for  the  engine 
or  the  amount  of  compressed  air  can  be  computed. 

The  next  step  is  to  find  the  displacement  of  the  engine.  If 
Vf  is  known,  the  amount  of  air  per  minute  at  the  upper  pressure 
is  given  by 

7'2  =  ^  (53) 

If  V'2  is  known  this  can  be  used  to  find  V'\,  the  amount  of  air 
between  the  ends  of  the  expansion  and  compression  lines: 


-tf 

V'i  V'i 

Displacement  =  -~-  = 


If  there  is  leakage,  D  =  (56) 


AIR  COMPRESSORS 


147 


Now  all  the  above  determinations  are  for  complete  expansion 
and  compression.  If,  however,  there  is  incomplete  expansion  or 
compression  other  calculations  must  be  made. 

Let  -  be  the  proportional  value  cut-off  and  x  be  the  value  of 

compression  expressed  in  terms  of  the  displacement  and  assume 
that  there  is  a  clearance  of  ID.  The  work  is  found  as  the  area 
of  the  card  after  the  pressures  at  3  and  6  (Fig.  56)  are  found. 


FIG.  55. — Diagram  for  air  engine  FIG.  56. — Diagram  from  air  engine 
with  complete  expansion  and  com-  with  incomplete  expansion  and  com- 
pression, pression. 


=  p't 


Pi 


-i 


w    ,  . 

Work  per  minute  = 


I   J 

33000  X  D.H.P.      1 


(57) 
(58) 


mech.  eff. 


1  -A; 


-s]Z>- 


1  -k 


pj, 


l-k 


(59) 


In  this  equation  D  is  the  unknown  quantity  and  may  be  found 
for  any  given  power  required.  The  quantity  in  brackets  is  the 
mean  effective  pressure  of  the  card.  The  amount  of  free  air 


148  HEAT  ENGINEERING 

required  for  this  motor  is  ascertained  by  determining  the  differ- 
ence in  the  weights  of  air  at  2  and  5  and  then  reducing  this  to 
volume.  If  there  is  leakage,  D  is  made  D  •*-  /  in  finding  volume. 


= 

~ 


BTl 


P 


D 


x)D 


BT 


*'6!8 

(60) 


PC 

Tz  and  T5  are  not  definitely  known  on  account  of  the  Action 
of  the  cylinder  walls.  On  a  test  they  could  be  determined,  since 
the  temperature  at  the  beginning  of  compression  is  the  same  as 
that  of  the  exhaust  and  from  this  the  mass  in  the  clearance  space 
would  be  known.  From  the  air  supply  the  mass  entering  would 
be  known  and  consequently  the  total  mass  at  the  point  of  cut- 
off or  release.  Since  the  pressures  and  volumes  at  these  points 
are  known,  as  well  as  the  masses,  the  temperatures  could  be 
found.  Although  not  strictly  correct,  T$  will  be  assumed  the 
same  as  T3  and  T2  will  be  assumed  equal  to  TV  This  gives 


(61) 


With  either  of  the  cases  above  if  D  is  known  the  stroke  and 
diameter  can  be  found  as  in  the  case  of  the  compressor. 
Assume  2LN  =  300  to  700  or  1000. 
Assume  N  and  find  L. 
Now  D  =  (Fh  +  Fe)LN. 
From  this  Fh  and  Fc  may  be  found. 

The  output  of  the  air  motor  is  usually  given  in  a  problem  and 
from  this  the  i.h.p.  may  be  found. 

=  i.h.p.  (62) 


.  ... 

mech.  eff. 

From  this  the  size  of  the  motor  and  the  amount  of  free  air  may 
be  found  after  the  pressure  limits  and  events  of  the  stroke  are 
assumed.  After  this  is  accomplished  the  drop  in  pressure  in 
the  supply  line  is  found  and  finally  the  pressures,  free  air  for  the 


AIR  COMPRESSORS  149 

compressor,  size  of  compressor  and  the  power  to  drive  the  same. 
The  overall  efficiency  is  found  by 

Overall  eff.  =  ^        h.p.  output  «rf  eogSne  (63) 

h.p.  required  to  drive  compressor 

This  efficiency  is  found  to  be  about  40  per  cent.  when  worked  out, 
although  with  leaks  a  lower  efficiency  is  obtained.  To  show  vari- 
ous values  of  efficiency  a  number  of  problems  will  be  worked  out 
later. 

FAN  BLOWERS 

The  turbo  compressors  and  fan  blowers  not  only  give  a  com- 
pression of  the  air  but  in  addition  the  velocity  of  the  air  at 
discharge  is  so  great  that  there  is  an  additional  term  for  the  gain 
of  kinetic  energy.  In  the  turbo  compressors  the  cooling  is 
practically  continuous,  although  it  may  be  considered  as  an  ra- 
stage  compressor  of  a  value  of  n  of  almost  unity  on  account  of 
the  cooling  effect  of  the  metal  and  the  water  jacket.  In  this 
case  the  expression  for  work  is 


(«, 


For  the  fan  blower  the  action  is  so  rapid  and  the  path  so  short 
that  the  action  is  assumed  adiabatic  and  the  expression  is 


In  this  pz  and  pi  differ  by  a  small  quantity. 

GOVERNING 

The  fan  and  turbo  compressor  are  of  value  because  the  quantity 
of  air  may  be  varied  with  the  need,  the  pressure  changing  with 
the  quantity  and  the  power  changing  within  known  limits  so  that 
motors  may  be  provided.  With  displacement  compressors  work- 
ing at  a  fixed  pressure,  the  simple  way  of  changing  the  quantity  is  to 
change  the  speed  of  the  compressor.  This  may  be  accomplished 
with  steam  engine  drives  by  a  slight  throttling  of  the  steam. 
The  point  of  cut-off  is  not  changed  and  the  pressure  is  only  slightly 
changed  because  for  a  given  delivery  pressure  for  the  air  a  definite 
steam  pressure  is  required  to  give  sufficient  area  on  the  steam 
card.  The  slight  increase  of  pressure  will  overcome  the  friction 


150 


HEAT  ENGINEERING 


and  speed  up  the  machine.  When,  however,  the  speed  of  the 
motor  cannot  be  varied,  as  is  the  case  with  certain  electric  motors, 
the  varying  demand  for  air  must  be  met  in  other  ways.  Among 
the  ways  suggested  to  care  for  a  varying  quantity  of  air  the 
following  may  be  mentioned :  (a)  delayed  closing  of  inlet  valve, 
(6)  delayed  closing  of  discharge  valves,  (c)  throttling  air  on  suc- 
tion and  (d)  changing  clearance.  These  methods  are  all  under 
the  control  of  the  governor.  All  except  method  (c)  mean  no 
change  in  efficiency.  Fig.  57  shows  methods  (a),  (6)  and  (c) 


Clearance 


\     Suction  Valve  Held  Open 
for  a  Certain  Time 


Ckara 


Discharge  Valve  Held  Open 
for  a  Certain  Time 


Suction  Pipe 
Throttled 


Change  in  Clearance 


FIG.  57. — Methods  of  changing  the  discharge  from  an  air  compressor  of 

fixed  speed. 

as  well  as  three  cards  for  three  different  clearances.  The  clear- 
ance is  varied  by  automatically  connecting  different  chambers 
to  the  cylinder  heads  as  the  pressure  rises.  This  is  controlled  by 
the  governor  as  the  pressure  rises,  necessitating  a  reduction  in  the 
quantity  of  air;  the  increase  of  clearance  will  cut  down  the 
quantity.  When  compressors  are  to  be  operated  at  different 
pressures,  the  work  of  the  steam  cylinder  is  controlled  by  a 
throttle  governor,  a  variable  cut-off  governor  or  by  a  Meyer 
valve  gear. 


AIR  COMPRESSORS 
MOTORS 


151 


The  engines  using  compressed  air  may  be  of  the  regular  form 
of  engine  or  its  equivalent.  Fig.  58  shows  the  section  through  a 
Little  David  riveter  of  the  Ingersoll  Rand  Co.  In  this  the 
throttle  valve  A  is  controlled  by  the  handle  B  which  is  pressed 
by  the  thumb  of  the  operator  allowing  air  to  enter.  The  valve  C 


FIG.  58. — Little  David  riveter  of  Ingersoll-Rand  Co. 

allows  air  to  enter  behind  the  piston  D  which  is  driven  at  a  high 
speed  against  the  shank  E  which  drives  the  rivet.  The  valve  C 
shown  in  black  admits  air  to  the  groove  at  /  and  exhaust  takes 
place  through  the  passage  running  from  the  left  end  of  the  piston. 
After  passing  through  passages  in  the  valve  the  exhaust  leaves 
through  the  outlet  to  the  left  of  C.  When  G  is  uncovered  air 


FIG.  59. — Ingersoll-Rand  air  drill  (Little  David). 

rushes  through  a  small  port  and  actuates  the  valve  so  that  air  is 
cut  off  from  /  and  the  port  H  is  connected  to  outlet  and  the  air 
to  the  right  of  the  piston  is  discharged.  A  small  port  at  this  time 
discharges  air  into  the  large  exhaust  passage  which  is  now  cut  off 
from  the  atmosphere  and  this  forces  the  piston  to  the  right. 


152 


HEAT  ENGINEERING 


When  the  piston  passes  H  the  air  remaining  to  the  right  and  an 
amount  which  constantly  leaks  through  a  passage  to  the  right  of 
/  is  compressed,  cushioning  the  piston.  The  increase  of  pressure 
here,  and  that  due  to  a  leak  into  the  groove  /  after  this  is  covered 
by  the  piston,  causes  the  valve  to  reverse  and  the  action  to  be 
repeated. 

Fig.  59  shows  the  construction  of  a  drill  which  is  the  equivalent 
of  four  single-acting  engines  with  rotary  valves  worked  from  the 
shaft  while  Fig.  60  shows  a  section  through  a  rotary  reamer  or 
drill.  This  is  a  rotary  engine.  Air  is  admitted  at  A  and  causes 
the  diaphragm  or  plate  B  to  turn 
on  the  axis  C. 

In  Fig.  61  a  preheater  is  shown. 
In  this  the  passage  of  the  air 
through  the  hollow  portions  of  the 
heating  surface  raises  the  tempera- 
ture to  a  high  point.  The  fuel 
may  be  coal,  oil  or  gas. 

The  arrangement  and  amount  of 


FIG.  60. — Rotary  air  drill. 


FIG.  61. — Sullivan  air  preheater. 


surface  in  the  intercooler  is  fixed  by  the  principles  of  Chapter 
III,  and  will  be  discussed  later  in  connection  with  a  definite 
design. 

LOSSES  IN  TRANSMISSION 

The  various  losses  discussed  are  shown  in  Fig.  62. 

abed  =  unavoidable  loss  due  to  two  staging.     This  may 

be  eliminated  if  a  two-stage  engine  is  used. 
efgd  =  loss  due  to  leakage  in  compressor. 
gfh  =  loss  due  to  change  of  line. 


AIR  COMPRESSORS 


153 


hfji  =  loss  due  to  cooling. 
ijkl  =  loss  due  to  leakage  in  line. 
knop  —  mnql  =  loss  due  to  throttling. 

mrst  =  loss  due  to  high  back  pressure. 


FIG.  62. — Diagram  showing  various  losses  in  air  transmission. 

LOGARITHMIC  DIAGRAMS 

Before  solving  problems  it  will  be  well  to  consider  the  con- 
struction of  polytropics  on  logarithmic  coordinates.  If  a  curve 
of  the  form  pVn  —  const,  be  plotted  on  the  coordinates  log  p  and 
log  V  it  is  found  that  the  equation  above  takes  the  form 

log  p  +  n  log  V  =  const. 

This  is  of  the  form  x  +  ny  =  const.,  or  the  curve  becomes  a 
straight  line  and  the  value  of  n  is  the  slope  of  the  curve  since 

dx 
dy  = 

If  the  coordinates  of  p  and  V  of  a  curve  are  plotted  logarithmic- 
ally with  logarithmic  coordinates  and  the  points  lie  in  a  straight 
line,  as  in  Fig.  63,  the  curve  must  be  of  the  form  pVn  =  const, 
and  the  slope  of  the  line  is  the  value  of  n.  The  value  of  n  in 
the  figure  is  to  be 

n  =  1.4 

In  this  figure  the  value  of  the  logarithm  extends  from  0  to  1 
and  the  numbers  from   1  to   10,  but  the  figure  is  constructed 


154 


HEAT  ENGINEERING 


beyond  that  limit  since  the  same  variation  in  logarithm  changes 
the  numbers  from  10  to  100,  100  to  1000,  etc.,  or  1  to  0.1, 
0.1  to  0.01,  etc. 


LogP 


idou 

0.9 

1 

\ 

I? 

0.8 

\ 

\ 

8.0 
80 
0.7 

\ 

\, 

C  0 

\ 

\ 

00.0 

coo 

0.6 

\ 

\ 

\ 

\ 

60.0 
600 

\ 

\ 

\ 

0.4 

\ 

\ 

N. 

140.0 
400 

V 

\ 

\ 

\ 

0.3 

\ 

\ 

\ 

30.0 
800 

d 

\ 

\ 

\ 

\ 

1 

\ 

0.2 

\ 

\ 

\ 

1 

\ 

\ 

\ 

v 

\ 

x 

\ 

\ 

\ 

0.1 

1.0 

\ 

\ 

\ 

lib0-1                                            C  0.2                      0.8                    0.4            0.6         0.6       0.7      0.8    0.9      I 
1.0                                            2.00                    S.O                  4-0            6.0         6.0      7.0     8.0    9.0     1 

10                                                 20                       30                     40             60           60        70      80     90    1C 

Log  V 


Table  from 

Log  Diag. 

P=  10  F=iOO 

p  .  20  V  61 

P  .  40  V"  37 

P  =  60  V  28 

P  =  100  V"  19 


V"  10 


1-20 


FIG.  63. — Logarithmic  diagram  with  repeated  coordinates  for  plotting 

pV1-4  =  const. 

If  in  the  figure  from  b  at  the  end  of  ab,  the  vertical  to  c  is  drawn 
and  then  cd  is  drawn  parallel  to  ab,  and  after  taking  d  to  e  and 
drawing  ef  with  one  following  fg,  a  series  of  lines  may  be  put  on  a 
single  figure  which  would  require  a  much  larger  figure  if  a  single 


AIR  COMPRESSORS 


155 


line  were  drawn  as  in  Fig.  64.     Fig.  64  is  less  confusing  than  Fig. 
63,  but  it  requires  much  more  space. 

Such  figures  as  the  above  are  of  value  in  constructing  curves 
of  expansion.  If  on  a  logarithmic  diagram  through  the  points 
log  pi  and  log  Fi,  a  line  inclined  at  n  =  1.35  be  drawn,  the  values 


1000 
800 
600 

400 
200 

100 
80 
60 

40 
7 
20 

10 

0 

\ 

\ 

\ 

\ 

\ 

\ 

\ 

\ 

\ 

\ 

\ 

\ 

N 

\ 

> 

\ 

\ 

\ 

\ 

\ 

\ 
\ 

.1        0,2         0.4    0.6   0.8  1.0        2.0        4.0    6.0  8.0  1C 
Log  V 

FIG.  64. — Logarithmic  diagram  with  continued  coordinates. 

of  p  and  V  for  any  point  may  be  found  and  the  pV  curve  drawn 
with  simplicity. 

To  aid  in  computations  of  many  of  the  above  quantities  charts 
and  diagrams  have  been  devised  from  which  results  may  be 
obtained  rapidly.  A  set  of  diagrams  arranged  by  Professor  C.  R. 
Richards  and  Mr.  J.  A.  Dent,  based  on  temperature  entropy 
logarithmic  diagrams,  is  of  great  value.  See  Bulletin  63  of  the 
Engineering  Experiment  Station,  University  of  Illinois. 

PROBLEMS 

To  apply  the  above  formulae,  a  problem  will  be  assumed  as  follows: 
Ten  5-h.p.  air  motors  are  to  operate  at  200  r.p.m.  between  60.3  Ibs.  gauge 
pressure  and  0.3  Ib.  gauge  back  pressure.  Find  the  size  of  the  motors  if  the 


156 


HEAT  ENGINEERING 


piston  speed  is  200  ft.  per  minute  and  find  the  amount  of  free  air  required 
to  drive  these  assuming  a  3  per  cent,  leakage.  Assume  first  that  the  com- 
pression and  expansion  are  complete  with  a  5  per  cent,  clearance  and  sec- 
ond assume  a  cut-off  at  0.35  stroke  and  compression  at  0.1  stroke  with 
5  per  cent,  clearance.  Compression  of  this  air  is  to  take  place  in  a  two-stage 
air  compressor  from  —  0.2  Ib.  gauge  pressure  to  135.3  Ibs.  gauge  pressure. 
The  compressor  is  3000  ft.  from  the  motors  and  the  drop  in  the  supply 
main  is  to  be  4  Ibs.  Find  the  size  of  the  main.  There  are  fifty  H2-in. 
holes  in  the  main;  find  the  leakage.  The  air  temperature  is  70°  F.  and  the 
available  water  is  at  60°  F.  The  leakage  in  the  compressors  is  3  per  cent. 
Find  the  air  to  be  taken  into  the  compressor  to  cover  all  leakages.  Find 
the  various  efficiencies,  losses  and  constants  of  the  system.  Would  it  be 
advisable  to  preheat  the  air  to  350°  F.?  In  this  problem  the  work  will  be 
done  by  the  use  of  a  12-in.  slide  rule  of  Log.  Log.  form. 


FIG.  65. — Indicator  card  for  air 
engine  with  complete  expansion 
and  compression. 


FIG.  66. — Indicator  card  for 
an  engine  with  incomplete  ex- 
pansion and  compression. 


Problem  1. — Indicated  horse-power  of  one  motor. 

I  Lh-P-  =  Tff7.=  qk=6-25 

Friction  1.25  h.p. 

Problem  2. — Displacement  and  free  air  for  two  cases, 
(a)  Complete  expansion  and  compression. 


6.25  X  33,000  =  Y~I 14-7  X  144  X  °'97 

r 
=  3.5  X  2118  X  0.977/ 1  1  - 


0.4 

L4|(40) 


78  cu.  ft.  per  minute 
14.7  X  78 


75 
=  15.26  X   (^ 


=  15.26 


V, 

F'2 


Ki  =  1  +  0.05  -  0.05  X 
0.97  X  48 


48 


D  = 


0.892 


(53) 
(54) 

=  1.05  -  0.158  =  0.892    (30) 
52  cu.  ft.  per  minute.  (56) 


AIR  COMPRESSORS  157 


(6)  Incomplete  expansion  and  compression. 
1.4 


PA  =  75[°005"+'l5        =  75  X  d*6  =  19-41bs-Per  ^are  inch    (57) 

roos  +  o  n1-4 

P« =  15        n  n^  =  15  X  4.66  =  69.9  Ibs.  per  square  inch  (58) 


M.e.p.  =  0.35  X  75  -  0.9  X  15  + 

19.4  X  1.05  -  75  X  0.4  -  15  X  0.15  +  69.9  X  0.05 

~-  0.4 

=  26.25  —  13.5  +  21  =  33.75  Ibs.  per  square  inch 
6.25  X  33,000  =  D  X  144  X  33.75 

D  =  42.5  cu.  ft.  per  minute 
42  *) 


x  75  x  a4  ~  15  x  144 


I  O^X    '  ~l 

ir)    x  al5J 


=  2.89  I  30.0  -  3.3J  =  77.2  cu.  ft.  per  minute. 
If  leakage  is  considered 

*>'  -  £gf  -  79-5  cu-  ft- 

The  loss  in  power  due  to  leakage  is  ((Toy  —  6.25  j  =0.19  h.p. 

It  is  seen  that  although  the  displacement  is  larger  for  the  complete  expan- 
sion arrangement  the  amount  of  free  air  is  slightly  less.  The  difference 
between  the  two  cases  is  not  great. 

Problem  3.  —  Size  of  motors. 
(a)  and  (6).     N  =  200 


.'.  L  =  x  ft.  =  6  in. 
(a)  D  =  52     =  (Ah  +  Ac]  X  200  X 

If  the  areas  are  equal 

52 

A   =  200  sq'  ft' 
52  X  144 

200        =37.4sq.m. 

Assuming  a  1-in.  piston  rod  and  tail  rod 

Acyi.  =  37.4  +  0.7854  =  38.2  sq.  in. 
d  =  6.95  in.  =  7  in. 

Cylinder  7  hi.  X  6  in. 

(6)  D  =  42.5  cu.  ft.  =  2  X  200  X  %  X  A 

A  =  2QQ  X  144  =  30.6  sq.  in. 

Acyi.  =  30.6  +  0.7854  =  31.4  sq.  in. 
d  =  6.33  in. 

Cylinder  6.33  in.  X  6  in. 


158  HEAT  ENGINEERING 

Problem  4. — Leakage  from  pipe  line. 
Pressure  at  one  end  =  135.3  +  14.7  =  150 
Assumed  pressure  for  leakage  =  150 


4   ~  32  X  32 
=  0.132  lbs.  per  second 
60  X  0.132  X  53.35X530 

V/*  =  "  14.7  X~144"~  =    °5  CU'  ft>  per  mmute- 

Problem  5. — Total  free  air  assuming  case  of  complete  expansion  motors, 
(a)  Air  for  motor  =     78  cu.  ft.  per  minute. 

(6)  Leakage  air  =  105  cu.  ft.  per  minute. 

(c)  Total  delivered  free  air          =  885  cu.  ft.  per  minute. 

O  Q  K 

(d)  Total  free  air  required  for  compressor  =  ^  Q  >  =  942  cu.  ft.  per  minute. 

(0.94  is  leakage  factor  for  two  stages.) 
Problem  6. — Diameter  of  pipe  line  to  carry  air. 
, .       885  X  14.7  X  144 

M  =  60  X  53.35  X  530  =  L1°  lbs'  per  sec°nd* 
For  first  approximation  b  =  0.02 

/ISO2  -  1462\  ,  0.02  X  l.l2  X  53.35  X  530 

( 2 ) 144    =  "~  ^2 

2  X  32.2  X  yg  X  d5 

d6  in  inches  =  d&  in  feet  X  12s  =  1050 

d  =  4.02  in.;  use  4-in.  pipe. 
6  should  then  be  checked 

53.35  X  530 


X 


148  X  144 


16.7  ft.  per  second. 


-  °-0124 


X  0.7854 

0.0274       0.00145  0.012 


=  0.0124  +  0.0016  +  0.0044  +  0.0020 

=  0.0204 

The  value  of  0.02  is  close. 
Using  equation  (46) 

8852  X  3000 


D6  =  990 

D  =  3.97  in.  or  4  in. 

The  drop  in  pressure  will  not  be  4  lbs.  if  a  4-in.  pipe  is  used  with  6  = 
0.0204 

'  X  3000  (44) 


p22  =  22,500  -  1238  =  21,262 
p2  =  145.8  lbs. 
.  '  .  Drop  =  4.2  lbs. 
If  a  one-third  increase  in  this  is  assumed  a  drop  of  6  lbs.  will  be  cared  for. 


AIR  COMPRESSORS  159 

Problem  7.  —  Loss  due  to  throttling  from  150  Ibs.  to  144  Ibs.  abs. 

0.4  O4 

Loss  -  £  H.7  X  144  X  *•[(£)  "-  ©  "]  (48) 

=  6,550,000[0.524  -  0.518] 

=  39,300  ft.-lbs.  per  min.  =  1.19  h.p. 

Problem  8.  —  Loss  due  to  throttling  from  150  Ibs.  to  75  Ibs.  abs. 

04 

Loss  =  6,550,000  [0^  IA  -  0.518  1  (48) 

=  6,550,000[0.631  -  0.518] 

=  6,550,000[0.113] 

=  740,000  ft.-lbs.  per  min.  =  22.5  h.p. 

Loss  if  leakage  occurs  at  higher  pressure 

780 
=  22.5  X  gg£  =  19.8  h.p. 

Problem  9.  —  Loss  due  to  cooling  from  T2  to  TV 

0.35 

53° 


1 

=  530(10.33)  7.7 
=  530  X  1.355  =  718°  F.  abs.  =  258°  F. 

0.35  04 

^X0.94X942XU.7X144[  @  «  ><  ™  -l]  [l- 

=  6,550,000[1.355  -  1][1  -  0.518] 

=  1,122,000  ft.-lbs.  per  min.  =  34.1  h.p. 

The  use  of  values  previously  computed  reduces  the  amount  of  computation. 
Problem  10.  —  Work  on  two-stage  compression  and  intermediate  pressure. 

0.35 


=  15,120,000[1  -  1.355] 

=  -  5,370,000  ft.-lbs.  per  min.  =  162.7  h.p. 

0  94  l   Q  97 
0.955  =  -  —  g  —  ~  ~~  =  leakage  factor  for  work  on  two 

stages 
p'z  =  V  150  X  14.5  =  46.6  Ibs.  per  sq.  in.  abs.    (19) 

Problem  11.  —  Work  on  single  stage. 

0.35 


=  7,470,000[1  -  1.833] 

=  6,220,000  ft.-lbs.  per  min.  =  188.5  h.p. 

Problem  12.  —  Saving  due  to  double  staging. 

Gain  =  188.5  -  162.7  =  25.8  h.p. 


160  HEAT  ENGINEERING 

Problem  13. — Saving  due  to  jacket  if  on  single  stage  (16). 

OA  0.35 

0»-  147XH4X  ^{£[l  -  gg)  "]  -  t»[l  -  (0)  '"]  j 

=  1,930, 000 {3.5[1  -  1.95]  -  3.86[1  -  1.833]} 

=  1,930,000  {  -  3.32  +  3.218) 

=  -  197,000  ft.-lbs.  per  min.  =  5.98  h.p. 

Problem  14. — Saving  due  to  jacket  if  on  two-stage  compressor. 

Gain  =  1,950,000{7[1  -  (1.95)^]- 7.72[1- (1.833)]) 
=  126,500  ft.-lbs.  per  min.  ~  3.84  h.p. 

Problem  15. — Unavoidable  loss  in  two  staging  (36) . 

0.35  0.35 

1.35  x     ."-^  ,    885  r        /150\-2TTr      /14.5\-2T~| 

Loss  =  035  X  14'7  X  144  X  0955 L1      \14^j        J  L1  ~  ( ISO/        J 

=  -  7,560,000[0.355][l  -  j-|^l 

=  —  705,000  ft.-lbs.  per  min.  =  21.25  h.p. 

Problem  16. — Loss  due  to  change  in  expansion  line  of  engine  from  n  to  . 
starting  from  point  at  end  of  compression. 

n-l  k-l 


=  14.7  X  144  X  885  X  1.355  1  3.86  j"l  -  37^33] 


=  2,540,000(1.755  -  1.706) 

=  124,500  ft.-lbs.  per  min.  =  3.78  h.p. 

Problem  17.  —  Loss  due  to  high  back  pressure  in  engine. 


=          X  780  X  14.7  X  144  [0.632  -  0.625] 
=  40,500  ft.-lbs.  per  min.  =  1.23  h.p. 
Problem  18.  —  Loss  due  to  leakage  in  compressor. 

Loss  =  available    work   without    leakage  —  available 

work  with  leakage 
=  [work    of    compression—  unavoidable    loss]  X 

[1  —  leakage  factor] 
=  [162.7  -  21.25][1  -  0.955]  =  6.38  h.p. 


AIR  COMPRESSORS 


161 


Problem  19.  —  Loss  due  to  leakage  in  line. 

Loss  =  [work  of  compression—  unavoidable  loss—  loss 
due  to  compressor  leakage—  loss  due  to  change 
..     .leakage  volume 
of  exponent-loss  due  to  cooling]     total  volume 

=  [162.7  -  21.25  -  6.38  -  3.78  -  34.1]^ 
=  11.5  h.p. 
Problem  20.  —  Power  to  drive  compressor. 

ind.  power       162.7 

Power  =   mech    eff.    =  O^  =  18°'8  h*- 
Friction  loss  =  180.8  -  162.7  =  18.1  h.p. 
Problem  21.  —  Power  of  motor  for  compressor. 


Motor  loss  =  200.1  -  180.8  =  19.3  h.p. 
TABLE  OF  POWER 


H.p. 

Percentage 

Power  supplied  motors  

200.1 

100.00 

Loss  motors  

19.3 

9.6 

Power  supplied  compressor 

180  8 

Loss  compressor  

18.1 

9.0 

Power  supplied  air.   .    . 

162  7 

Unavoidable  loss  

21.25 

10.6 

Loss  from  leakage  in  compressor. 

6  38 

3.2 

Loss  due  to  change  from  n  to  k  
Loss  due  to  cooling  after  delivery. 

3.78 
34.1 



1.9 
17.1 

Available  power  supplied  line 

97  2 

Loss  due  to  leakage  in  line  

11.5 

5.7 

Loss  due  to,  throttling  to  75  Ibs  
Available  power  at  engine 

19.8 

65  9 

9.9 

Loss  due  to  high  back  pressure  . 

1  23 

0  6 

Loss  due  to  leakage  3  per  cent,  of  65  9 

1  98 

1  0 

Available  indicated  work  

62.70 

Amount  loss  in  engine  20  per  cent. 

12  54 

6  3 

Amount  delivered  

50.20 

25.1 

Saving  due  to  jacket.  . 

3  84 

100.00 
1  9 

Saving  by  double  staging 

25  8 

12  9 

It  is  to  be  noted  that  there  is  9.9  per  cent,  loss  due  to  throttling  and  5.7 
per  cent,  due  to  leakage.  These  might  be  saved,  giving  the  overall  effi- 
ciency about  40  per  cent,  in  place  of  25  per  cent.  The  low  efficiency  has 
been  due  to  the  leakage  from  line  and  throttling.  In  good  installations 
40  per  cent,  overall  efficiency  might  be  reached, 
ll 


162  HEAT  ENGINEERING 

Problem  22.  —  Heat  removed  by  jacket  (15). 


14-7H 


X 

0.35 


=  -  1.73  X     j    X  13,830  X  144  X  0.355 

=  —  436,000  ft.-lbs.  per  min. 
=  —  561  B.t.u.  per  min. 

Problem  23.  —  Heat  removed  by  intercooler  (28). 

0.35 


=  +  2,400,000  ft.-lbs.  per  min. 
=  3090  B.t.u.  per  min. 

Problem  24.  —  Water  for  jacket. 

Assume  water  rises  from  75°  to  90°  F.  in  jacket. 

P\A  1 

G  =  QQ  __  75  =  37.4  Ibs.  per  min. 

Problem  26.  —  Water  for  intercooler. 

Assume  rise  in  temperature  from  60°  to  75°  F. 

3090 

G  =  7K  _  fiQ  =  206  Ibs.  per  min. 

Problem  26.  —  Mean  At,  assuming  counter  current  flow  from  60°  to  75° 
F.  for  water  and  70°  to  258°  F.  for  air,  using  Rensselaer  formula. 

Aii  =  258  -  75  =  183°  F. 
A£2  =  70  -  60  =  10°  F. 


Note  that  this  is  quite  different  from  ~^~-  =  97°  F. 

Problem  27.  —  Surface  required  and  arrangement. 
Assume  water  velocity  =    3  ft.  per  sec. 
Assume     air     velocity  =  10  ft.  per  sec. 

46:6X144X1         (10  -  1.75)°  •«.    oW 
Value  of  K  -  126  X  53.35(164  +  460)  X  -  -^-  -X  3*  =  16.7 

(44)  Ch.  I 

,  3090  X  60 

Surface  =  x  66  =  169  sq.ft. 


Arrangement  of  ^-in.  tubes  12  ft.  long  with  water  inside. 
Area  of  1  tube  =        X      X  12  =  2.35  sq.  ft. 


AIR  COMPRESSORS  163 


169 
No.  of  tubes  =     ~oc  =  ^  tubes. 


No.  of  tubes  in  one  nest. 

Area  of  inside  of  tube  *|  X0  ^JJ/'^X  jpj-  "  °-00258  scl-  ft- 

Amount  of  water  =  206  Ibs.  per  min. 

Velocity  of  water  =  3  ft.  per  sec.,  180  ft.  per  min. 

OOA 

No.  tubes  in  parallel  =  64.5  x  180  x  Q.00258  =  7  tubes' 

This  means  10  nests,  or  the  water  will  run  from.  7  tubes  to  7  tubes 
so  as  to  give  a  path  of  120  ft.  in  cooler. 

14  7        614 
Volume  of  air  per  min.  =  (942  X  0.97)  ^g  X  ^Q  =  340  cu.  ft. 

Area  for  air  =  in  y  gn  =  0.565  sq.  ft. 

Suppose  the  7  tubes  are  made  in  a  case  so  that  the  air  may  pass  along. 
Area  of  passage  0.565  sq.  ft.  =  81  sq.  in.  To  this  the  area  of  seven  H-m. 
tubes  or  3  sq.  in.  will  be  added  giving  84  sq.  in.  for  passage  of  air. 

Problem  28.  —  Size  of  compressors  at  80  r.p.m.  with  400  ft.  piston  speed. 


K,-  1+0.05-  0.05 

=  0.932 

885 


Low-pressure  displacement  =  nn/i  \x  n  noo  =  1010  cu-  ^-  Per  mm- 

u.yT:  /s.  u.yoz 

1010    =   400  Amean 

Anet  =  2.52  sq.  ft.  =  388  sq.  in. 
Assume  a  3-in  piston  rod;  A  =  7.06  sq.  in. 
Agross  =  363  sq.  in. 
dcyi.  =  21.5  in. 

T  40Q  9W  ff 

L  =  2~><80- 

Size  of  compressor  21.5  in.  X  30  in. 
High  pressure  displacement. 


Intermediate  pressure    =  \/150  X  14.5  =  46.6  Ibs. 


Stroke  =  30  in. 

OAQ 

Area  net  =  ^  =  0.77  sq.  ft.  =  111.0  sq.  in. 

Area  gross  =  111.0  +  7.06  =  118.06 

dcyi.  =  12.3  in. 

Size  of  high-pressure  compressor  12.3  in.  X  30  in. 
Problem  29. — The  efficiency  of  preheating  is 

0.4 

Eff.  =  1  -   (11^  1A  =  1  -  0.631  =  0.369 


With  the  leakage  and  throttling  giving  an  efficiency  of  25  per  cent,  on 
output  or    (25  +  6.3)  =  31.3  per  cent,  on  indicated  work  of  engine  it  is 


164  HEAT  ENGINEERING 

seen  that  this  would  pay.     If  there  had  been  no  throttling  and  the  air 
were  at  a  150-lb.  absolute  pressure  the  efficiency  would  be 

q.4 

Eff.  =  1  -   frJLl IA  =  1  -  0.518  =  0.482 


This,  being  better  than  40  per  cent.,  would  pay.  In  most  cases  pre- 
heating will  pay.  It  is  well  to  note  in  passing  that  the  efficiency  of  pre- 
heating depends  on  the  range  of  pressure. 

TOPICS 

Topic  1. — What  are  the  different  methods  of  compressing  air?  Sketch 
machines  for  this  purpose.  Under  what  conditions  is  each  used? 

Topic  2. — Explain  the  construction  of  a  two-stage  air  compressor  and  give 
the  reasons  for  the  use  of  such  apparatus.  What  is  the  purpose  of  the  inter- 
cooler?  Why  is  moisture  apt  to  collect  in  the  air  space  of  the  intercooler? 
How  is  it  removed  ?  What  is  the  peculiar  form  of  inlet  valve  used  on  the  low- 
pressure  piston? 

Topic  3. — For  what  reason  is  air  for  a  compressor  taken  from  the  atmos- 
phere and  not  from  the  engine  room?  Why  are  water-jackets  used?  Ex- 
plain the  action  of  the  Taylor  hydraulic  air  compressor. 

Topic  4. — Derive  the  expression: 


Topic  6.  —  What  is  the  effect  of  clearance?  Prove  this.  Derive  the  ex- 
pression for  the  clearance  factor. 

Topic  6.  —  Explain  what  is  meant  by  volumetric  efficiency.  Derive  the 
formula  for  true  volumetric  efficiency  in  terms  of  the  leakage  and  clearance. 
Show  how  to  find  the  horse-power  to  drive  a  compressor. 

Topic  6a.  —  Why  is  jacketing  of  value?  Is  this  true  under  all  conditions? 
Is  jacketing  very  effective?  Derive  the  expressions  for  the  heat  removed  by 
the  jacket,  the  saving  by  the  jacket  and  the  water  required  for  the  jacket. 

Topic  7.  —  What  is  multistaging?  Sketch  a  figure  and  show  why  this  is  of 
value.  What  is  the  function  of  the  intercooler?  How  large  should  it  be? 
Derive  the  expression  for  two-stage  compression: 


Topic  8.  —  Using  expression  of  Topic  7,  show  that  the  work  is  a  minimum 
when 

p'z  =  Vpipz' 

What  are  the  conditions  for  minimum  work  for  a  three-stage  compressor? 
Reduce  the  expression  for  work  to  a  simple  form.  To  what  does  the 
expression  reduce  for  an  ra-stage  compressor? 

Topic  9.  —  Derive  the  formula  for  the  temperature  at  the  end  of  compres- 
sion on  the  line  pVn  =  const,  when  pi,  pz,  and  T\  are  known.  What  is  the 
value  of  the  intermediate  temperatures  on  a  multistage  compressor?  Show 
that  these  are  the  same  on  each  stage. 

Topic  10.  —  Derive  the  formula  for  the  heat  removed  by  the  intercooler  and 
that  for  the  amount  of  water  required?  What  is  the  effect  of  leakage  in 
multistage  compression? 


AIR  COMPRESSORS  165 

Topic  11. — Explain  method  of  finding  displacement  of  the  various  cylin- 
ders of  a  multistage  compressor  to  deliver  a  given  amount  of  air. 

Topic  12. — Derive  formulae  for  the  moisture  removed  from  the  intercooler 
and  for  the  condition  of  the  air  discharged  from  it.  How  are  the  inlet  and 
outlet  pipes  and  valves  designed? 

Topic  13. — Derive  a  formula  for  the  saving  due  to  multistaging.  Derive 
the  expression  for  the  unavoidable  loss  due  to  two-stage  compression. 

Topic  14. — Derive  a  formula  for  the  work  done  in  a  single-stage  engine. 
What  is  the  temperature  at  the  end  of  expansion?  To  what  does  the  expres- 
sion for  work  reduce  for  a  two-stage  engine? 

Topic  16. — Derive  an  expression  for  the  loss  due  to  cooling  after  compression. 

Topic  16. — Derive  a  formula  for  the  loss  due  to  a  difference  in  the  lines  of 
expansion  in  the  engine  and  of  compression  in  the  compressor. 

Topic  17. — Derive  the  formula  for  the  loss  of  pressure  in  a  pipe  line  carry- 
ing air.  Why  is  T  assumed  constant?  What  is  the  effect  of  leakage? 
How  is  the  amount  of  leakage  determined? 

Topic  18. — Derive  the  expression  for  the  loss  due  to  throttling  and  one  for 
the  gain  from  preheating. 

Topic  19. — Derive  the  expressions  for  the  displacement  of  an  air  engine 
with  complete  expansion  and  compression  and  with  incomplete  expansion 
and  compression  to  give  a  certain  power. 

Topic  20. — Derive  the  expression  for  the  work  of  a  fan  blower.  Explain 
how  the  quantity  of  air  from  a  constant-speed  compressor  is  regulated. 

Topic  21. — Explain  by  means  of  a  diagram  the  various  losses  which  enter 
into  an  air-transmission  system.  Give  the  efficiency  in  terms  of  the  ratio 
of  two  areas. 

Topic  22. — Explain  the  use  of  logarithmic  diagrams  in  plotting  curves  of 
the  form  pVn  =  const.  Explain  the  method  of  increasing  the  range  of 
a  diagram. 

Topic  23. — Sketch  the  forms  of  some  of  the  air  motors  in  common  use  and 
explain  their  action. 

PROBLEMS 

Problem  1. — Find  the  power  to  compress  1000  cu.  ft.  of  free  air  per  min- 
ute from  14.5  Ibs.  absolute  to  60  Ibs.  gauge  in  a  single-stage  air  compressor; 
n  =  1.35,  leakage  =  3  per  cent.,  clearance  =  2  per  cent. 

Problem  2. — Find  the  power  to  compress  1000  cu.  ft.  of  free  air  per  minute 
from  —0.3  Ibs.  gauge  pressure  to  160  Ibs.  gauge  pressure  in  a  two-stage  air 
compressor  with  n  =  1.35,  leakage  3  per  cent,  in  each  cylinder,  clearance  3 
per  cent.  Find  the  intermediate  pressure. 

Problem  3. — Find  the  size  of  the  compressors  in  Problems  1  and  2  if  they 
operate  at  120  r.p.m.  and  are  double-acting. 

Problem  4. — Find  the  temperature  at  the  end  of  compression  in  Problems 
1  and  2  if  the  original  temperature  were  75°  F. 

Problem  5. — Find  the  heat  removed  in  the  intercooler  for  compressor  of 
Problem  2.  Find  the  amount  of  water  required  if  the  water  available  is  at 
60°  F.  and  is  arranged  by  counter  current  to  leave  at  80°F.  Find  AT7 
assuming  that  the  value  of  the  coefficient  of  heat  transmission  is  constant. 
Find  the  value  of  K  by  assuming  the  velocity  of  the  air  to  be  30  ft.  per  second 
and  the  velocity  of  the  water  as  10  ft.  per  second  and  using  Nicolsou's 
formula. 


166  HEAT  ENGINEERING 

Problem  6. — Using  the  variable  parts  of  the  expression  for  work  find  the 
relative  work  for  single,  two-stage  and  three-stage  compression  between  14.5 
Ibs.  and  290  Ibs.  absolute.  Find  the  clearance  factor  for  these  assuming  3 
per  cent,  clearance  in  each  cylinder.  Find  the  temperature  at  the  end  of 
compression  in  each  case  assuming  75°  F.  as  the  original  temperature. 

Problem  7. — Find  the  size  air  main  to  deliver  3000  cu.  ft.  of  free  air  per 
minute  when  compressed  to  145  Ibs.  absolute,  if  the  temperature  is  75°  F. 
and  the  drop  in  pressure  allowable  is  2  Ibs.  in  160  ft. 

Problem  8. — Find  the  efficiency  of  transmission  if  1000  cu.  ft.  of  free  air 
per  minute  is  compressed  to  190  Ibs.  gauge  pressure,  transmitted  800  ft. 
in  a  tight  main  and  used  after  storage  in  engines  at  150  Ibs.  pressure  in  a 
single-stage  engine.  Assume  all  constants  needed.  Is  the  result  the 
probable  efficiency  or  the  maximum  efficiency? 

Problem  9. — Find  the  various  losses  in  the  system  described  in  Problem  8. 

Problem  10. — Three  H-in.  holes  are  in  a  pipe  line  carrying  1500  cu.  ft.  of 
free  air  per  minute  compressed  to  125  Ibs.  gauge  pressure  and  at  80°  F.  What 
is  the  percentage  loss  due  to  these  holes? 

Problem  11.— The  wet  bulb  reads  70°  F.  in  80°  F.  weather  when  the  ba- 
rometer stands  at  29.85  in.  This  air  is  compressed  to  50  Ibs.  gauge  pressure 
in  the  intercooler  at  80°  F.  Has  any  moisture  been  precipitated?  If  not 
what  is  the  relative  humidity  of  the  air?  If  precipitation  occurs  how  much 
precipitation  occurs  per  minute  if  1000  cu.  ft.  of  free  air  are  handled  per 
minute? 

Problem  12. — Find  the  size  of  an  air  engine  to  develop  15  h.p.  (a)  with 
complete  expansion  and  compression,  and  (6)  with  cut-off  at  30  per  cent., 
compression  20  per  cent.,  clearance  5  per  cent.  p\  =  725  Ibs.  absolute,  pb  = 
14.8  Ibs.  absolute.  R.p.m.  =  125. 

Problem  13. — Air  at  1500  Ibs.  absolute  pressure  is  stored  in  a  tank  and  when 
used  in  an  engine  it  is  throttled  to  150  Ibs.  absolute  pressure.  The  tempera- 
ture in  the  tank  is  80°  F.  What  is  the  temperature  at  the  end  of  complete 
expansion  in  the  engine  to  1 5  Ibs.  absolute  pressure  ?  How  much  of  the  energy 
in  the  tank  is  available  in  the  engine  after  throttling? 

Problem  14. — In  Problem  13  the  air  on  leaving  the  throttle  valve  is  heated 
by  a  vapor  lamp  to  450°  F.  By  what  per  cent,  is  the  power  of  the  engine 
increased  by  this?  What  is  the  temperature  of  discharge  from  the  engine? 

Problem  15. — An  air  compressor  of  Problem  1  has  the  compression  line 
reduced  to  pv1'35  =  constant  by  the  jacket.  How  much  heat  does  the  jacket 
remove?  How  much  water  does  this  require  if  the  water  range  is  from  70°  F. 
to90°F.? 

Problem  16. — One  thousand  cubic  feet  of  free  air  per  minute  is  compressed 
from  15  Ibs.  pressure  to  150  Ibs.  pressure  (absolute)  on  two  stages.  The 
original  temperature  was  60°  F.  Find  the  final  temperature  of  discharge. 
Find  the  percentage  loss  if  this  air  is  allowed  to  cool  to  60°  F.  before  it  is 
used  in  the  air  motors;  n  =  1.35. 

Problem  17. — How  much  loss  occurs  in  Problem  15  due  to  the  fact  that  the 
expansion  in  the  engine  is  on  the  line  piP  =  const,  instead  of  pF1-35  = 
const.  ?  Express  this  as  a  percentage. 

Problem  18. — Find  the  values  of  p  and  v  at  several  points  on  the  line, 
p  yi.25  _  const.,  passing  through  the  point  p  =  165  Ibs.  V  =  3  cu.  ft.  by 
means  of  a  logarithmic  diagram.  Use  pressures  of  150,  100,  50  and  25  Ibs. 


CHAPTER  V 
THE  STEAM  ENGINE 

On  account  of  the  numerous  improvements  made  upon  it 
during  the  two  centuries  of  its  use,  the  steam  engine  was,  until 
quite  recently,  the  most  important  heat  engine  using  steam. 
The  steam  turbine  has  taken  this  place  at  present  due  to  the 
decreased  cost  of  production,  the  greater  concentration  of  power, 
and  the  good  efficiencies  over  a  wider  range  of  load. 

The  engine  was  developed  without  any  theory  of  thermodynam- 
ics by  Worcester,  Papin,  Savery,  Newcomen  and  Watt,  and  to 
account  for  the  action  of  the  engine  in  later  times,  Rankine  in 
England  and  Clausius  in  Germany  developed  a  mathematical 
theory  of  heat.  The  later  development  of  the  engine  is  connected 
with  the  names  of  Corliss,  Porter,  Reynolds,  Willans,  Sultzer 
and  Stumpf.  The  engine  has  been  improved  by  increasing  the 
Carnot  efficiency  through  the  use  of.  higher  initial  pressures  and 
superheat  and  by  lowering  the  back  pressure,  while  the  use  of 
jackets,  reheaters,  superheated  steam,  four  valves,  heavy  lagging, 
multiple  staging  and  flow  in  one  direction  have  all  had  their 
effects  on  its  practical  efficiency. 

In  the  steam  engine  (Fig.  67)  steam  containing  heat  is  ad- 
mitted to  the  cylinder  A  by  the  valve  B,  so  that  it  moves 
the  piston.  The  valve  B  is  operated  by  the  valve  rod  C 
from  an  eccentric  or  crank  D  so  that  when  the  piston  E  has 
reached  a  certain  point  steam  is  cut  off  from  the  cylinder  and 
the  steam  is  allowed  to  expand  to  a  point  near  or  at  the 
end  of  the  stroke  (release  point)  when  the  valve  is  so  moved 
as  to  allow  the  steam  to  escape  to  the  outside.  The  steam 
admitted  to  the  other  end  or  the  inertia  of  the  fly  wheel  causes  the 
piston  to  return  in  its  stroke  forcing  the  steam  out  of  the  cylinder 
until  a  point  near  the  end  of  the  stroke  (point  of  compression)  is 
reached  when  the  exhaust  is  closed  and  the  movement  of  the 
piston  compresses  the  steam  until  the  valve  again  connects  this 
side  of  the  cylinder  t<\  the  steam  chest  F  and  steam  pipe  G  so 
that  steam  enters  once  more.  In  theory  the  indicator  card 

167 


168 


HEAT  ENGINEERING 


THE  STEAM  ENGINE 


169 


170 


HEAT  ENGINEERING 


will  appear  as  in  Fig.  68.  The  expansion  and  compression  are 
complete.  The  point  of  cut-off  is  at  1,  release  at  2,  compres- 
sion at  3  and  admission  at  4.  The  quality  of  the  steam  at  3 
is  the  same  as  that  at  2,  hence  the  condition  at  4  is  the  same 
as  that  at  1  if  the  lines  1-2  and  3-4  are  assumed  to  be  adiabatics. 
3-4  is  the  compression  line  of  the  steam  in  the  clearance  space 
4-5.  If  these  are  adiabatics  with  MQ  pounds  of  steam  on  3-4 
and  M  +  M0  pounds  of  steam  on  1-2,  Fig.  69,  with  no  clearance 
space  and  with  the  weight  of  steam  M  on  the  line  1-2  between 
the  same  pressures  and  qualities  at  the  limiting  points  will 
have  the  same  area  and  represent  the  cycle  with  no  clear- 
ance. This  is  true  because  in  theory  the  clearance  steam  ex- 


FIG.  68. — Indicator  card  from  en- 
gine with  complete  expansion  and  com- 
plete compression. 


FIG.  69. — Indicator  card  with 
no  clearance. 


pands  and  contracts  along  the  same  line  requiring  and  giving  no 
work.  M0  is  called  the  clearance  steam  and  M  is  called  the 
working  steam,  steam  supply  or  boiler  steam. 

CYCLE  OF  STEAM  ENGINE 

Fig.  69  then  represents  the  theoretical  form  of  cycle  used  on 
the  steam  engine.  It  is  known  as  the  Rankine  or  Clausius 
cycle  although  the  former  name  is  employed  at  times  for  the 
cycle  shown  in  Fig.  71.  The  difference  between  the  Carnot 
cycle  and  the  Rankine  cycle  is  that  in  the  former  cycle  the 
working  substance  is  supposed  to  be  in  the  cylinder  and  heat 
only  is  added,  while  in  the  latter  substance  is  added  and  with  it 
heat.  On  the  Carnot  cycle,  if  represented  by  Fig.  68,  the  quality 
changes  from  that  at  4  to  that  at  1,  growing  greater,  while  it 
decreases  from  2  to  3,  hence  1-2  and  4-3  are  not  similar  adia- 
batics. For  the  Rankine  cycle  the  quality  on  4-1  is  constant, 
the  variable  being  the  mass  and  this  is  true  on  2-3.  In  both 


THE  STEAM  ENGINE'  171 

cases  4-1  and  2-3  are  isothermals  if  saturated  steam  is  used. 
The  heat  added  on  the  Carnot  cycle  is  Mrdx  and  on  the  other 
is  xrdm,  the  total  amount  being  the  same  in  each. 

In  Fig.  69  the  exhaust  steam  could  be  used  to  heat  an  equal 
weight  of  water  to  the  temperature  of  exhaust  for  boiler  feed, 
since  this  could  be  done  by  the  heat  of  the  liquid  of  the  steam 
even  if  the  water  supply  were  at  32°  F.  With  the  heat  of  vapori- 
zation in  addition  for  the  vapor  part  of  the  mixture,  this  and 
even  more  could  be  done.  This  additional  amount  is  of  no  value 
for  the  operation  of  the  engine  although  for  warming  other 
water  it  has  a  great  value.  The  point  which  must  be  grasped 
by  the  student  is  the  fact  that  the  heat  in  the  exhaust  could,  if 
properly  used,  heat  the  feed  necessary  for  the  working  steam  to 
a  temperature  of  the  exhaust  pressure.  For  this  reason,  the 
datum  plane  from  which  heat  is  to  be  measured  is  always  taken 
as  at  the  temperature  of  the  exhaust. 

HEAT  AND  EFFICIENCY  OF  CYCLE 

The  heats  added  on  the  different  lines  are: 

M(ii  -  q'o)  on  4-1  (1) 

0  on  1-2 

Mfa  -  q'o)  on  2-3  (2) 

0  on  3-4,  since  M  =  0 

(Note.  —  q'0  is  used  rather  than  q'%  because  0  is  used  to  refer 
to  the  datum  in  all  cases.) 


.'.AW  =  Qi  -  Q2 

It  is  to  be  remembered  that  i\  and  z'2  are  the  heat  contents  at 
two  points  on  the  same  adiabatic  and  hence  they  are  found  in 
the  same  entropy  column  or  line. 

Qi  =  M[ii  -  qf0] 

Theoretical  eff.  =  r?3  =  ?*  ~  *'  (5) 

i\  —  q  o 

This  is  sometimes  called  the  Rankine  efficiency. 

In  some  texts  the  work  of  pumping  the  water  into  the  boiler 
from  4  to  1  is  considered  giving  net  work 

ii  -  i2  -  A  (pi  -  pz)v' 


172 


HEAT  ENGINEERING 


The  last  term  is  small.     It  really  does  not  belong  to  the  engine 
being  one  of  the  charges  against  the  boiler. 

Now,  the  Carnot  eff.  =  771  =  — ^ — 


If  M  pounds  of  steam  are  required  per  horse-power  hour,  the 
work  =  2546  B.t.u.  and  the  heat  is  M(ii  —  q'0). 

2546 


(6) 


175 


FIG.  7Q.—T-S  diagram  of 
Rankine  cycle. 


FIG.  71. — Rankine  cycle  with  incom- 
plete expansion. 


The  T-S  diagram  for  this  is  an  aid   in  studying   the  cycle. 
It  is  shown  in  Fig.  70.     In  this,  points  3  and  4  come  together. 
The  heat  added  per  pound  from  3  to  1  is 
6-4-3-1-5  =  ii  -  q'0 

Heat  12  =  0 

Heat  23  =  6-3-2-5  =  i2  -  q'0 

Work  =  64315-6325  =  3412  =  ^  -  t2 
4-a-l-2  would  show  the  Carnot  cycle. 

If  the  expansion  is  not  carried  to  the  back  pressure  line  it  is 
called  incomplete  expansion  and  the  cycle  using  such  is  then 
sometimes  spoken  of  as  the  Rankine  cycle.  The  best  way  to 
study  this  cycle  is  by  dividing  it  into  two  parts  by  the  line  2-6, 
Fig.  71. 

Area  5126  =  M(il  -  za) 

Area  2346  =  M[A(p2  -  p3>2] 

AW  =  451234  =  M[(ii  -  i2)  +  A(pz  -  ps)v2]  (7) 


THE  STEAM  ENGINE  173 

J!  -  12  +  A  (p2  -  p3)v2 

"3  =  ~  t!  -  q'o  (8) 

!Ti-  To 

171  • 


2546 
775  ~  Mfe  -  q'o) 


773  for  the  engine  cycle  is  sometimes  called  the  Rankine  efficiency 
of  the  cycle  or  this  term  is  applied  to  equation  (5). 

In  the  above  equations,  i\  and  i%  are  heat  contents  on  the 
same  adiabatic,  hence,  if  the  entropy  is  known  for  point  1,  i*  is 
found  at  pressure  2  in  same  entropy  column  as  i\  if  there  is  a 
table  or  on  the  same  entropy  line  from  a  chart.  If  there  is 
neither  table  nor  chart  which  can  be  used,  the  formula  for  the 
adiabatic  is  to  be  used  to  find  the  quality  at  2  and  from  this,  zV 

Fi  =  Mxiv'i  (To  find  xi). 


«'i  +     r  =  **  +         (To  find  xj 

F2  =  Mx2v"z  (To  find  F2). 

ii  =  q'i  +  Wi  (To  find  z'i). 

^2  =  q'z  +  ^2^2  (To  find  z"2). 

All  necessary  terms  are  known  for  any  of  these  formulae  if 
taken  in  the  order  given.  The  value  of  x\  being  theoretical  and 
therefore  the  highest  possible,  is  that  of  the  boiler  steam  and 
should  be  so  used. 

STEAM  CONSUMPTION 

From  equations  (3)  and  (7)  the  amount  of  steam  per  horse- 
power hour  theoretically  required  is  given  by 

«=s 

™  2546 

M    =  -.  --  r—        -7-r-  „  (10) 

ti  -  ^2  +  A  (p2  -  pi)  v"i 

To  illustrate  the  applications  of  the  formulae,  suppose  an 
engine  uses  26  Ibs.  of  steam  per  horse-power  hour  at  110  Ibs. 
gauge  pressure  with  x  =  0.98  and  has  a  pressure  at  the  end 
of  expansion  of  15  Ibs.  gauge  and  a  back  pressure  of  0.5  Ibs. 
gauge.  It  is  desired  to  find  the  various  efficiencies. 


174  HEAT  ENGINEERING 

From  Peabody's  Temperature          From  Marks  and  Davis7 

Entropy  Tables  Charts 

pi  =  110  +  14.7  =  124.7  pi  =  124.7 

ti  =  344.2  si  =  1.563 

81  =  1.563  ^  =  1173 

11  =  1172.5  p72  =  29.7 
p2  =  15  +  14.7  =  29.7  s2  =  1.563 
s2  =  1.563  i2  =  1065 

12  =  1065.8  z2  =  0.897 
v2  =  12.55  v2  =  12.5 

?'0  of  (0.5  +  14.7)  Ibs.  =  181.9 
t0  =  213.7 

By  computation  the  following  results,   using  the  tables  of 
Marks  and  Davis: 

p1  =  129.7 

i2  =  318.4  +  0.98  X  872.7  =  1173.4 

81  =  0.4996  +  0.98  X  1.0820  =  1.56 

p2  =  29.7 

S2  =  1.56  =  0.3673  +  x    X  1.3326 

1.1927 

X*  =  E3326  = 

i2  =  218.2  +  0.895  X  945.6  =  1065 
v2  =  0.895  X  13.89  =  12.42 
q'0  =  q'  of  (14.7  +  0.5)  =  181.6 

Any  of  the  three  sets  of  values  could  be  used.     Using  the 
first  ^         „  ±9 

344.2  -  213.7       130.5 

771  =344.2  +  459.6  =  803^  =  16'2  per  Cent 

%.   f       A  x 

1172.5  -  1065.8  +  ^(29.7  -  15.2)  144  X  12.55 
77o     T» 

173  =  1172.5  -  181.9 

139.9 
=  QQQ-^  =  14.2  per  cent. 

14.2 

772  =  T^~2  =  87.2  per  cent. 

2546 2546  _ 

776  ~  26(1172.5-181.9)  ~  2580  ~ 

9.9 

174  =  .^7-n  =  70  per  cent. 


THE  STEAM  ENGINE 


175 


The  results  of  the  above  computations  show  that  the  efficiency 
of  the  theoretical  cycle  is  more  than  87  per  cent,  of  that  of  the 
most  perfect  cycle  but  that  the  actual  engine  only  utilizes  70 
per  cent,  of  the  amount  which  should  be  available  theoretically. 
For  these  reasons  there  is  not  much  chance  of  improving  the 
cycle  of  the  steam  engine  but  to  make  174,  the  practical  efficiency, 
greater  the  cylinder  is  covered  with  some  non-conducting  material, 
a  steam  jacket  is  used  or  superheated  steam  may  be  employed. 
These  all  tend  to  cut  down  the  transfer  of  heat  to  and  from  the 
cylinder  walls  and  thus  cut  down  the  amount  of  steam  required. 

TEMPERATURE— ENTROPY  DIAGRAMS 

The  cycle  is  shown  on  the  T-s  diagram,  Fig.  72.  The 
free  expansion  from  2  to  3  cuts  off  the  corner  of  the  figure  as 
shown  by  2-3.  This  figure  and  Fig.  70  are  used  to  represent 


FIG.  72. — T-S  diagram  of  Rankine 
cycle  with  incomplete  expansion. 


8'8  7  S 

FIG.  73. — Effect  of  increasing  tem- 
perature range. 


the  Rankine  cycle  without  and  with  complete  expansion.  They 
are  not  true  representations  since  the  lines  of  Figs.  70  and  72 
represent  additions  of  heat  and  not  of  substance  as  is  the  case 
on  the  actual  cycle.  Because  the  heat  added  in  the  figures  on 
the  various  lines  represents  the  heat  added  on  the  lines  with  the 
substance  the  figures  are  used  to  represent  this  cycle.  More- 
over, the  area  of  the  figure  1234  of  Fig.  70  or  123456  of  Fig.  72 
represents  the  difference  between  Qi  and  Qz  and  hence  repre- 
sents the  work. 

The  efficiency  is  represented  by  the  area  of  the  cycle  divided 
by  the  total  area  beneath  4-1  in  Fig.  70  or  72  and  hence,  if  the 
temperature  is  increased,  both  of  these  areas  become  greater  and 
the  efficiency  is  increased.  This  is  shown  by  6i-l"  of  Fig.  73. 

461l//234       461234 
Jjjtl*  ~   846il'7  >  84617 


176 


HEAT  ENGINEERING 


If,  now,  the  lower  temperature  is  lowered  from  3  to  3',  the 
denominator  of  the  fraction  is  changed  a  slight  amount,  8'4'48, 
while  the  numerator  is  changed  by  the  area  433 '4'.  In  this  way  the 
efficiency  is  increased  in  most  cases.  If  the  free  expansion  is  so 
great,  as  shown  in  Fig.  73,  that  the  distance  3-4  is  small,  it  maybe 
that  the  added  area,  433'4',  may  not  be  greater  than  8'4'48 
and  the  efficiency  is  not  increased  by  lowering  the  back  pressure. 
As  the  free  expansion  2-3  becomes  less,  as  2'-3"  (which  means 
that  the  expansion  1-2  is  more  nearly  complete),  the  line  3"-4 
is  so  long  that  the  gain  in  efficiency  due  to  a  decrease  in  back 
pressure  is  great.  The  increase  in  pressure  to  give  a  perceptible 
increase  in  temperature  for  the  boiler  steam  is  so  great  that  the 

pressure  must  be  increased  considerably 
to  raise  the  line  6-1  an  appreciable 
amount.  Hence  the  gain  in  efficiency 
from  an  increase  of  pressure  is  not  as 
marked  as  a  slight  decrease  in  the  lower 
pressure  of  the  cycle  when  the  tempera- 
ture change  is  more  rapid.  Thus,  to  de- 
crease the  back  pressure  from  15  Ibs.  to 
2  Ibs.  abs.  means  a  change  of  87°  F. 
while  an  increase  from  100  Ibs.  to  113 
Ibs.  means  an  increase  of  10°  F.,  from 
150  Ibs.  to  163  Ibs..  8°  F.,  while  from  200 
Ibs.  to  213  Ibs.  means  an  increase  of  less 

than  6°  F.  It  must  also  be  remembered  that  a  decrease  in  the 
back  pressure  with  much  free  expansion  does  not  necessarily 
give  an  increase  of  efficiency. 

In  order  to  increase  the  efficiency  by  increasing  the  upper 
temperature  without  an  increase  in  pressure  superheated  steam 
is  used.  Theoretically  it  will  be  seen  that  although  this  does 
increase  the  efficiency  there  is  not  much  gain  because  this  heat 
is  not  added  at  constant  temperature.  Fig.  74  illustrates 
this  case 

Q1  =  9c618  =  q\  -  q'0  +  r  +  fcpdt  =  n  -  q'0 
Q2  =  94328  =  i<,  -  q'0  -  A(p2  -  pjv* 
Qi  -  Qz  =  work  =  45c6  1234 

Now  the  triangle  l-d-b  is  the  part  of  the  figure  which  repre- 
sents the  increase  due  to  superheat  and  being  added  to  5  cbd  234 


FIG.  74.— T-S  diagram 
of  Rankine  cycle  with 
superheated  steam. 


THE  STEAM  ENGINE 


111 


as  well  as  to  9  c  b  d  8  it  increases  the  efficiency,  having  the  greater 
effect  on  the  small  numerator.     This  reasoning  is  made  evident 


by  observing  that  the  efficiency 


5  cbd  234 


is  about  the  same 


95  cbd  8 

as  the  efficiency  of  the  cycle  of  Fig.  71  with  saturated  steam  and 
the  same  pressure  ranges  and  the  expression  for  the  latter  is 
made 'into  that  for  Fig.  74  by  adding  b-l-d  to  numerator  and 
denominator. 

EFFECTS  OF  CHANGES 

These  facts  have  been  shown  by  the  late  Prof.  H.  W.  Spangler 
in  his  Applied  Thermodynamics.  The  curves  of  Figs.  75,  76 
and  77  show  how  the  theoretical  efficiency  is  increased  as  the 
pressures  and  superheats  are  changed. 


•op 

30  Jl 

2U/ 

0 

11 

40|i 

280^ 

40^ 

X 

gao^ 

^ 

'     ^ 

oudens 
acuum 

ing 

28^" 

M 

IIOK 

0^° 

7 

^  Theoretical  Effic 

s  s 

o  •**.  v». 

„—  —  - 

'       Non-Coud 

Busing 

_       — 

_  —  ^ 

•^ 

1 

»*    125*  150^    175#  20 
Gage  Pressure 

0"     74"     15"     22  4"  3 
Inches  Vacuum 

)°       50°      100°    150°  20< 
Degrees  of  Superheat 

FIG.  75.  FIG.  76.  FIG.  77. 

FIG.  75. — Effect  of  change  of  steam  pressure  on  efficiency,  complete  ex- 
pansion and  saturated  steam. 

FIG.  76. — Effect  of  change  of  back  pressure  on  efficiency  (150  Ib.  steam 
1°  superheat),  complete  expansion. 

FIG.  77. — Effect  of  change  of  superheat  on  efficiency  (151  Ib.  steam  to 
28.5  in  vac.),  complete  expansion. 


These  figures  are  all  drawn  for  complete  expansion  and,  al- 
though in  all  cases  the  effect  of  vacuum  increase  is  greater  than 
the  effect  of  the  same  change  in  the  initial  pressure,  the  effect 
is  not  so  marked  as  that  shown  in  Fig.  76  when  there  is  incom- 
plete expansion.  These  figures  are  all  drawn  from  theoretical 
considerations. 

Although  the  effect  of  superheat  is  theoretically  slight,  its 
effect  on  the  actual  efficiency  may  be  much  more  pronounced 
since  it  cuts  down  certain  losses  which  occur  in  the  cylinder  and 
therefore  its  practical  effect  is  very  valuable. 
12 


178  HEAT  ENGINEERING 

TESTS  OF  ENGINES 

To  actually  determine  the  steam  consumption  of  an  engine  a 
test  is  made  while  the  load  is  kept  constant  on  the  engine.  If 
possible  the  exhaust  steam  is  condensed  in  a  surface  condenser 
and  weighed  at  regular  intervals  and  at  the  same  time  indicator 
cards  are  taken  and  readings  are  made  on  the  scales  .of  the 
Prony  brake,  or  watt  meter  if  the  engine  is  used  to  drive  a  genera- 
tor, on  the  calorimeter  to  determine  the  quality  of  the  steam, 
on  the  pressure  gauge  and  barometer,  on  the  temperature  and 
pressure  of  the  exhaust  steam,  on  the  temperature  of  the  con- 
densate  and  on  the  revolution  counter.  From  this  data,  as  will 
be  shown,  the  steam  per  horse -power  hour,  the  pressure  and 
quality  of  the  steam  supply,  and  the  pressure  of  the  exhaust 
may  be  found  to  be  used  for  the  actual  thermal  efficiency. 

CALORIMETERS 

The  calorimeter  used  to  determine  the  quality  of  the  steam 
may  be  of  three  forms:  the  electric,  the  separating  and  the 
throttling  calorimeters.  The  latter  is  shown  in  Fig.  78.  The  form 
here  shown  is  the  Barrus  Universal  Throttling  Calorimeter,  so 
called  because  by  the  addition  of  the  drip  pot  C  it  may  care  for 
steam  of  any  moisture  content.  In  this  a  sample  of  steam  is 
taken  preferably  from  a  vertical  steam  pipe  A.  In  horizontal 
pipes  it  has  been  found  that  much  of  the  moisture  separates  out 
and  flows  along  the  bottom  of  the  pipe  so  that  a  fair  sample 
cannot  be  obtained.  The  sampling  tube  B  is  a  pipe  closed  at 
the  end  and  containing  a  number  of  small  holes  around  its  circum- 
ference and  along  its  length  to  take  steam  from  various  parts  of 
the  pipe.  This  steam  is  carried  into  the  drip  pot  C,  which  is  in 
reality  a  separator,  where  most  of  the  moisture  is  taken  from 
the  steam.  The  height  of  the  water  in  the  drip  pot  is  shown 
by  the  glass  gauge  D  and  the  pressure  is  shown  by  the  gauge  E. 
Steam  then  passes  over  through  a  pipe  and  around  a  thermometer 
well  containing  the  upper  thermometer  F  and  from  here  it 
passes  through  a  small  hole  in  the  throttle  plate  G  to  the  ther- 
mometer well  containing  the  lower  thermometer  H  and  from  here 
to  the  atmosphere.  The  mercury  gauge  /  shows  the  pressure 
around  the  lower  thermometer  well.  The  throttle  plate  is  held 
between  flanges  but  insulated  from  them  by  some  non-conducting 
material. 


THE  STEAM  ENGINE 


179 


To  find  the  amount  of  water  collected  in  the  drip  pot  a  string 
is  tied  around  the  glass  gauge  and  when  the  water  reaches  that 
level  the  time  is  noted.  The  water  in  the  drip  pot  is  drawn  off 
at  any  convenient  time  into  a  cup  of  cold  water  J  so  that  the 
water  level  falls  below  the  string.  This  is  done  by  opening  the 
valve  K  and  putting  the  end  of  the  drip  pipe  beneath  the  water 
level  in  the  cup  so  that  none  of  the  hot  water  from  the  drip  pot 


FIG.  78. — Barrus  universal  calorimeter. 

can  evaporate  on  reaching  a  place  of  atmospheric  pressure. 
The  cup  is  lowered  after  closing  K  and  after  catching  all  the 
drip  its  change  of  weight  will  give  the  weight  of  water  drawn 
off.  The  time  at  which  the  water  again  reaches  the  level  of  the 
string  is  noted  and  the  interval  beween  the  two  times  of  passing 
the  mark  will  give  the  time  taken  for  this  water  to  be  caught. 
This  is  reduced  to  pounds  per  minute  or  second.  Call  this 
m  pounds. 


180  HEAT  ENGINEERING 

To  find  the  weight  of  steam  passed  through  the  throttle  hole 
in  the  same  interval  of  time,  this  steam  is  exhausted  into  water 
and  condensed  or,  what  is  quite  customary,  the  hole  is  measured 
and  Napier's  formula 


is  used. 

If  the  separator  were  perfect  the  quality  of  the  steam  coming 
from  the  pipe  would  be, 


since  M  is  the  weight  of  dry  steam  in  M  -f-  in  pounds  of 
mixture.  This  drip  pot,  if  perfect,  would  be  a  form  of  separat- 
ing calorimeter.  This  is  not  a  perfect  separator  and  so  the 
quality  of  the  steam  leaving  is  determined  by  throttling  the 
steam. 

The  steam  is  throttled  from  the  pressure  shown  by  E  to 
that  shown  by  /.  Throttling  action  means  constant  heat 
content.  On  expanding  steam  which  is  practically  dry,  it 
becomes  superheated. 

ie  =  it 
q'e  +  xere  =  q'i  +  r{  +  fcpdt  =  it  (12) 


The  thermometer  H  gives  the  number  of  degrees  superheat 
when  the  saturation  temperature  corresponding  to  the  pressure 
at  I  is  found.  Since  the  thermometer  may  register  incorrectly 
due  to  the  fact  that  the  mercury  column  projects  beyond  the 
top  of  the  thermometer  well,  the  thermometer  F  is  used  to  check 
this  and  determine  the  error.  The  thermometer  wells  are  filled 
with  heavy  oil  or  with  mercury  and  the  thermometers  are  inter- 
changed after  each  reading  so  as  to  equalize  errors  in  the  ther- 
mometers. After  this  is  done  the  temperature  of  saturation 
at  the  average  pressure  shown  by  E  is  found  and  the  difference 
between  this  and  the  average  reading  of  F  is  the  error  due  to 
stem  exposure.  An  error  proportional  to  the  amount  of  exposed 
mercury  column  is  assumed  and  the  average  reading  of  H  is 
corrected.  From  this  the  degrees  of  superheat  are  found,  then 


THE  STEAM  ENGINE  181 

ii  and,  after  this,  xe  is  found  from  equation  (13).     The  weight 
of  dry  steam  is  then 

xM 

xM 

and  the  actual  x  is  X  =  -^—  '-.  (14) 

M  H-  m 

Part  of  this  X  is  due  to  radiation  from  the  instrument  and  to 
find  this  correction  the  instrument  is  operated  with  steam  slightly 
superheated  or  dry  steam  and  the  apparent  quality,  xdry,  is 
found. 


is  the  correction  to  be  applied  for  this  instrumental  condensa- 
tion. A  common  way  to  run  this  dry  test,  as  it  is  called,  is  to 
attach  the  calorimeter  to  a  boiler  with  banked  fires  and  a  shut 
stop  valve  in  which  case  the  steam  in  the  steam  space  is  dry. 

Thermometers  are  sometimes  calibrated  to  read  correctly  when 
immersed  to  a  certain  depth  and  if  such  are  used  there  is  no  need 
for  the  upper  thermometer.  Such  thermometers  are  called 
constant  immersion  thermometers. 

Suppose  the  averages  of  a  number  of  readings  taken  at  five- 
minute  intervals  are  as  given  below  and  the  quality  is  desired. 

Pi  =  112-lb.  gauge 

Pz  =2  in.  Hg. 

Error  in  gauge  =  +  2  Ibs. 

Barometer  =29.8  in. 

Ti  =  335.20°  F. 

T2  =  265.2°  F. 

Wt.  of  drip  =  0.2  Ibs.  per  10  min. 

Area  of  hole  =  0.01  sq.  in. 

Exposed  column  =  120°  F.  for  TL 

=  100°  F.  for  T2. 

Observed  pi  112  Ibs. 

Error  2 

Corrected  pi  110  Ibs. 

Barometer  =  14.64 

Absolute  pi  114.64 

Tl  sat.  337.92°  F. 

Ti  observed  335.2°    F. 

Error  Tl  =  2.72°  F. 


182  HEAT  ENGINEERING 

100 
Error  T2  =  2.72  X    ~       =  2.27°  F. 


Observed  T2  265.2°    F. 

Corrected  T2  =  267.47°  F. 

Observed  p2  =  2  in.  Hg.      =  0.98  Ibs. 

Barometer  =         14.64 
Absolute  p2  15.62  Ibs. 

Temp.  sat.  =       215.  Q7°  F. 

Corrected  T2  =       267.  47°  F. 
Degrees  of  superheat  52  .  40°  F. 

i  for  15.62  Ibs.  and  52.40°  F.  superheat  =  1177  B.t.u  (from 
Marks  &  Davis) 

q'  for  114.64  Ibs.  =  308.7 

r    for  114.64  Ibs.  880 

308.7  +  880  x  1177 


m  =  0.2  Ibs.  in  10  min. 

-    '      •     M  _  nun*  0.01  x  6Q  x  1Q  _  9  83 

0.986X9.83 
9.83  +  02^ 

The  object  of  the  drip  pot  is  to  care  for  any  large  amount  of 
moisture  for,  if  the  amount  of  moisture  is  over  4  per  cent.,  the 
equation 

q'  +  XT  =  ii 

would  give  an  ii  so  small  that  the  steam  could  not  be  super- 
heated and,  of  course,  the  moisture  in  the  low-pressure  steam 
could  not  be  known.  Hence  the  upper  x  could  not  be  found. 
The  maximum  amount  of  moisture  possible  with  steam  super- 
heated on  the  lower  side  of  the  orifice  depends  on  the  pressure 
but  4  per  cent,  is  the  usual  amount  for  pressures  used  in  practice. 
Wherever  the  thermometer  /  shows  212°  F.  the  drip  pot  is  needed 
as  there  is  too  much  moisture  present. 

The  electric  calorimeter  uses  an  electric  current  to  dry  the 
steam  and  by  calibrating  the  same  the  amount  of  power  for  each 
per  cent,  of  moisture  at  a  given  pressure  can  be  found  and  from 
this  the  quality  for  the  actual  power  is  determined. 


THE  STEAM  ENGINE  183 

A  condenser  is  not  always  applicable  for  the  determination 
of  the  weight  of  steam  supplied  to  an  engine  and  in  that  case 
the  engine  is  supplied  from  a  boiler  or  group  of  boilers  which  have 
been  blanked  off  from  the  others.  In  this  way  the  weight  of 
boiler  feed  gives  the  steam  supplied  to  the  engine  if  the  weight  of. 
water  left  in  the  boiler  at  all  times  is  constant.  This  condition 
is  approximated  by  keeping  the  level  of  the  water  in  the  boiler 
gauge  constant  but  even  in  such  a  case  the  change  of  temperature 
or  of  the  rapidity  of  firing  may  make  this  an  uncertain  guide. 
To  eliminate  the  effect  of  this  uncertainty  such  tests  should 
extend  over  a  considerable  time,  say  4  to  6  hours  and  in 
some  cases  24  hours.  If  a  condenser  is  used  a  test  of  an  hour  or 
less  is  sufficient  for  accurate  results  after  the  engine  is  brought 
to  a  uniform  condition. 

ANALYSES  OF  TESTS 

To  study  the  action  of  the  cylinder  walls  two  methods  of 
analyzing  these  test  results  will  be  examined,  one  analytical, 
the  other  graphical. 

Hirn's  Analysis. — Hirn's  analysis  of  the  cylinder  performance 
of  steam  engines  consists  in  operating  a  test  for  a  certain  length 
of  time  and  from  the  observations  computing  the  actual  perform- 
ance of  the  engine.  To  better  illustrate  this  analysis  a  case  will 
be  computed.  From  the  test  the  following  average  results  are 
found : 


Time  of  test 60  min. 

Size  of  engine 10  X  15 

Clearance . 10  per  cent. 

No.  of  revolutions  during  test 14,400 

Pounds  of  steam  used 2356 

Average  weight  of  steam  per  card,  o  y  1440Q     =    0.0816  Ibs. 

Average  pressure  at  throttle 105  Ibs. 

Barometric  pressure 14 . 5  Ibs. 

Average  of  quality  of  steam 0 . 99 

Average  temperature  of  condensateT 120°  F. 

Average  temperature  of  water  leaving  condenser!  105°  F. 

Average  temperature  of  water  entering  condenser!  85°  F. 

Weight  of  condensing  water  used 94,240  Ibs. 

Weight  of  water  per  pound  of  steam 40  Ibs. 


184 


HEAT  ENGINEERING 


Average  results  from  indicator  cards 

Point  of  admission  at  1 0.0  per  cent. 

Point  of  cut-off  at  2 25  per  cent. 

Point  of  release  at  3 90  per  cent. 

Point  of  compression  at  4 22  per  cent. 

Abs.  pressure  at  1 48    Ibs.  per  sq.  in. 

Abs.  pressure  at  2 107  Ibs.  per  sq.  in. 

Abs.  pressure  at  3 37  Ibs.  per  sq.  in. 

Abs.  pressure  at  4 16  Ibs.  per  sq.  in. 

Work  area  during  admission,  Wa  =  5-1-2-6 2.83  sq.  in. 

Work  area  during  expansion,  Wb  =  6-2-3-7 3.91  sq.  in. 

Work  area  during  exhaust,  We  =  (3-10-8-7)    - 

(9-4-10-8) -0.98  sq.  in. 

Work  area  during  compression,  Wd  =  5-1-4-9.. .  —0.52  sq.  in.  • 


596  78 

FIG.  79. — Average  indicator  card  from  test  for  Hirn's  analysis. 


The  results  above  have  all  been  averaged  from  the  various 
readings  and  cards.  The  head  end  and  crank  end  have  been 
averaged  together  although  at  times  they  are  worked  up  sepa- 
rately the  weight  of  steam  being  divided  between  the  two  ends 
in  proportion  to  the  total  volume  at  the  point  of  cut-off  although 
there  is  no  true  foundation  for  this  method.  Another  method 
of  division  may  be  used  as  shown  by  Clayton.  See  page  210. 

The  first  point  to  compute  in  Hirn's  analysis  is  the  weight  of 
clearance  steam  per  card,  M0.  At  the  point  4  steam  is  assumed 
to  be  dry.  This  has  been  indicated  by  experiments  as  far  as 
they  have  been  tried  although  these  experiments  were  difficult 


THE  STEAM  ENGINE 


185 


to  perform  and  are  not  so  reliable  as  would  be  desired.     With 
this  assumption 

74  =  My  4 


or 


M   - 

Mo  ~  v\ 

74  =  total  volume  at  4 
v'\  =  specific  volume  of  dry  steam  at  4 


(15) 


v"  4  is  found  from  the  steam  tables.     After  this  the  weight  of 
the  working  steam  per  card  is  found. 


M  = 


weight 


2  X  No.  of  revolutions 


(16) 


Knowing  M  and  M0  the  quality  of  the  steam  at  the  various 
events  of  the  cycle  may  be  found  if  the  pressures  and  volumes 
at  these  points  are  known  from  the  cards. 

7r 


M0v'\ 


(M 


7s 


(17) 
(18) 

*'  "  (M  +  MoK's  (19) 

From  the  pressures  the  values  of  the  heat  of  the  liquid  and 
the  internal  heat  of  vaporization  may  be  found  and  from  these 
the  intrinsic  energy  at  all  points  may  be  computed. 

AUi  =  M0(q'i  +  Xipi)  (20) 

AC/2  =  (M  +  M0)(qf2  +  XZPZ)  (21) 

AUs  =  (M  +  Mo)(q'3  +  XSPS)  (22) 

P4)  (23) 


From   the   indicator   cards   the   external   works   during   the 
various  periods  may  be  found. 

Work  of  admission  =  Wa  =  5126  (24) 

Work  of  expansion  =  Wb  =  6237  (25) 

Work    of    exhaust  =  Wc  =  -  94108  +  73108  (26) 

Work  of  compression  =  Wd  =  -  5149  (27) 

The  energy  added  to  the  engine  from  the  outside  during 
admission  is 

Oi  =  M(q'  +  XT)  (28) 


186  HEAT  ENGINEERING 

The  energy  removed  is 

Q2  =  -  M[q'0  +  G(q'd  -  q't)  (29) 

G    —  weight  of  condensing  water  per  pound  of  steam. 

M  =  weight  of  steam  per  card. 

q',  x,  and  r  are  for  conditions  of  steam  at  the  throttle  valve. 

q'0  =  heat  of  liquid  at  temperature  of  condensed  steam. 

q'd  —  heat  of  liquid  at  temperature  of  discharge  condensing 

water. 
q'i  =  heat  of  liquid  at  temperature  of  inlet  condensing  water. 

Now  the  heat  added  during  any  event  is  found  by 

Q  =  A[U2  -  U,  +  W]  (30) 

If  the  heat  coming  from  the  cylinder  walls  during  the  different 
events  is  represented  by  Qaj  Qb,  Qc  or  Qd,  using  the  equation 
above,  the  various  quantities  are  given  by  the  equations 

Ql   +   Qa    =   AU2   -   AVi   +  AWa  (31) 

Qa  =  -  Qi  +  AU*  -  AUi  +  AWa  (32) 

Qb  =  AU3  -  AU2  +  AWb  (33) 

Qc  =  -  Q2  +  AC/4  -  AU,  +  AWC  (34) 

Qd  =  AUi  -  AU*  +  AWd  (35) 

If  these  result  in  positive  quantities  heat  is  given  up  by  the 
walls  while  negative  values  mean  that  heat  is  given  to  the  cylinder 
walls.  If  these  are  added  together  the  net  amount  must  be 
equal  to  the  amount  given  or  taken  by  the  cylinder  walls.  If 
there  is  no  source  of  heat  in  the  cylinder  walls  they  could  not 
give  heat  so  that  the  sum  could  not  be  positive.  If  negative  this 
heat  would  finally  melt  the  iron  if  abstracted  for  a  considerable 
time.  Since  the  walls  do  not  change  in  temperature  this  negative 
sum  must  equal  the  heat  radiated  from  the  cylinder  and 

Qa   +   Qb  +   Qc   +   Qd   =    Qr  (36) 

A  check  equation  for  Qr  may  be  determined  by  blocking  the 
engine,  filling  the  cylinder  with  boiler  steam  and  measuring  the 
condensation.  Then 

Qr  =  Mr(q'  +  xr-  q'0)  (37) 

If  there  is  a  steam  jacket  there  could  be  a  positive  sum  as 
heat  could  then  be  given  up  by  the  walls  from  the  heat  in  the 
jacket.  In  this  case 

/    Qa  +  Qb  +  Qc  +  Qd  +  Qi  =  Qr_  (38) 


THE  STEAM  ENGINE 


187 


and  Qf  is  determined  by  finding  the  weight  of  steam  condensed 
in  the  jacket  while  the  engine  is  running.  M3-  is  the  condensa- 
tion due  to  the  heat  given  to  the  steam  within  and 


i  =   My  (</'  +  XT  -   q'0) 


(39) 


Qr  must  be  determined  while  the  engine  is  at  rest  by  the  con- 
densation in  the  jacket. 

This  analysis  will  give  the  quantities  Qa,  Qb,  Qc  and  Qd,  but  it 
will  not  give  the  value  of  any  of  them  for  a  portion  of  an  event. 
It  will  tell  that  so  much  heat  has  been  given  or  received  during 
an  event  but  it  will  not  indicate  at  what  point  this  or  any  part 
of  it  has  been  given  up.  If  the  value  of 


V  =  (M  +  M0)xl  v'\ 


(40) 


be  found  at  cut-off  and  for  several  other  pressures  the  saturation 
curve  may  be  drawn  on  Fig.  79.  The  ratio  of  the  volume  on 
the  expansion  line  to  that  on  this  line  for  any  pressure  will  give 
the  quality  x.  This  quality  is  sometimes  plotted  on  the  card  as 
shown  in  Fig.  79. 

These  equations  will  now  be  applied  to  the  test  data.  The 
engine  is  10  in.  in  diameter  and  of  15-in.  stroke.  The  volume 
swept  out  in  one  stroke  is  therefore  0.682  cu.  ft.  and  if  the 
card  is  4  in.  long  and  has  been  drawn  with  a  40-lb.  spring 
the  scale  of  area  is. 

40  X  144  '     0.682 
area  scale  =  — -=== —  X  — r —  =  1.260  B.t.u.  per  sq.  in» 

If  10  per  cent,  of  the  card  length  be  added  to  each  event  expressed 
in  per  cent,  of  the  stroke  to  give  the  total  percentage  volume, 
these  when  multiplied  by  the  displacement  will  give  the  total 
volume  in  cubic  feet  at  each  event.  A  table  is  then  made  for 
the  events  as  shown  below,  using  Peabody's  tables. 


Event 

Actual  volume 

Pressure 

Specific 
volume 

0.' 

P' 

1 

0.068 

48  Ibs. 

8.84 

247.8 

846.2 

2 

0.238 

107  Ibs. 

4.16 

303.5 

801.5 

3 

0.682 

37  Ibs. 

11.29 

231.6 

858.6 

4 

0.220 

16  Ibs. 

24.74 

184.6 

893.8 

188  HEAT  ENGINEERING 


0.0089  Ibs. 


0.068 

Xl  =  0.0089  X  8.84  = 
0.238 

~ 


z  ~  (0.0816+0.0089)4.16 

0  682 

X*  =  0.0905  X  11.29  =  °'668 
Ul  =  0.0089  [247.8  +  0.865  X  846.2]  =  8.72 
U*  =  0.0905  [303.5  +  0.633  X  801.5]  ==  73.50 
C/3  =  0.0905  [231.6  +  0.668  X  858.6]  =  72.80 
C/4  =  0.0089  [184.6  +  893.8]  =  9.60, 
/  AW  a  =  2.83  X  1.260  =  3.57  B.t.u* 
/    AWb  =  3.91  X  1.260  =  4.92  B.t.u. 

AW0  =  -  0.98  X  1.260  =  -  1.23  B.t.u. 
AWd  =  -  0.52  X  1.260  =  -  0.65  B.t.u. 

=  0.0816  [312.0  +  0.99  X  877.1]  =  96.5  B.t.u. 

=  -  0.0816  [88  +  40  (73.0  -  53.1)]  =  -  72.0  B.t.u. 

=  -  96.5  +  73.5  -  8.72  +  3.57  =  -  28.15  B.t.u. 

=  72.80  -  73.5  +  4.92  =  +  4.22  B.t.u. 

=  72.0  +  9.60  -  72.8  -  1.23  =  +  7.57  B.t.u. 

=  8.72  -  9.60  -  0.65  =  -  1.53  B.t.u. 


Qr  =  -  16,89 

Total  work 6.61  B.t.u. 

Heat  supplied  above  32°  F 96.5    B.t.u. 

Heat  supplied  above  temperature  of  exhaust 81 .3     B.t.u. 

Heat  removed  by  walls  during  events: 

/  Admission 28. 15  B.t.u. 

/  Expansion , —4.22  B.t.u. 

/  Exhaust -7.57  B.t.u. 

1  Compression 1  >53  B.t.u. 

TEMPERATURE  ENTROPY  ANALYSIS 

The  graphical  temperature  entropy  analysis  is  to  be  used 
next  and  in  this  the  same  test  is  made  but  the  only  observations 
necessary  are: 

Size  of  engine 10  in.  X  15  in. 

Average  pressure 119. 5  Ibs.  per  sq.  in.  abs. 

Average  quality 0 . 99 


THE  STEAM  ENGINE 


189 


Weight  of  steam 2356  Ibs. 

Revolutions 14,400 

Clearance 10  per  cent. 

Average  compression 22  per  cent. 

Average  indicator  card Fig.  81 

This  method  has  been  developed  by  a  number  of  persons. 
Prof.  Boulvin's  method  has  been  combined  with  that  of  Reeves 
in  reducing  the  method  given  below. 

The  average  indicator  card  is  constructed  by  averaging  the 
pressures  at  ten  or  more  equidistant  points  along  the  card  or  a 


Temperature       in        O     Cu.   Ft.        Volume  in  Cu.  Ft. 


Deg.   F 


FIG.  80.— LTemperature  entropy  diagram  for  analysis  of  engine  performance. 

card  may  be  selected  on  which  the  m.e.p.  is  equal  to  the  average 
m.e.p.  of  the  test.     M  and  M0  are  then  found. 

M0  =  —^-  =  0.0089  as  tefore. 

V  4 


M 


Wt 


2  X  rev. 


=  0.0816  Ibs.  as  before. 


The  chart  for  the  Temperature  Entropy  analysis  is  now  to  be 
drawn.  This  consists  of  four  quadrants,  Fig.  80,  a  p-v  quadrant 
having  a  line  for  1  Ib.  of  saturated  steam,  a  T-p  quadrant  having 
the  line  of  saturated  steam,  a  T-s  quadrant  having  the  liquid 
line  for  1  Ib.  of  water  and  the  saturation  line  for  1  Ib.  of  steam 


190 


HEAT  ENGINEERING 


and  a  V-s  quadrant  which  is  used  for  transferring  the  values  of 
x.  This  figure  is  constructed  by  aid  of  the  steam  tables  and 
takes  the  form  shown  in  Fig.  80. 

To  transfer  the  indicator  card  to  this  figure,  a  number  of 
points  on  the  mean  diagram,  Fig.  81,  are  taken  and  the  heights 
from  the  line  of  absolute  zero  pressure  and  lengths  from  absolute 
zero  of  volume  are  measured  and  tabulated.  The  lengths  are 
then  multiplied  by  the  scale  of  volume  of  Fig.  81  and  divided  by 
the  product  of  the  scale  of  Fig.  80  and  the  value  of  M  +  M0 
in  order  to  get  the  distance  to  each  point  if  1  Ib.  of  steam  were 
present  on  the  expansion  line.  The  heights  are  multiplied  by 
the  scale  of  81  and  divided  by  the  scale  of  80.  Thus,  in  the 


2    35 


FIG.  81. — Average  indicator  card  from  test  arranged  for  T-S  analysis. 


problem  considered  there  is  a  4-in.  card  for  a  10  X  15  engine 
with  a  40-lb.  spring  and  . 

M 

M  +  Mo  =  "b.0816  +  0.0089  =  0.0905 

Fig.  80  has  scales  of  2.5  cu.  ft.  to  the  inch  and  20  Ibs.  per  square 
inch  to  the  inch.  The  cylinder  has  a  displacement  of  0.682  cu.  ft. 
with  a  4-in.  card.  The  lengths  are  therefore  multiplied  by 

"IT  X  00905  X  2^  =  °'754 

in  order  to  get  the  length  to  lay  off  in  Fig.  80  so  as  to  represent 
the  card  for  1  Ib.  of  steam  on  the  expansion  line.  The  multi- 
plier for  height  is  merely  40/20  =  2.00.  These  distances  are 
tabulated  to  aid  in  the  work  as  shown. 


THE  STEAM  ENGINE 


191 


Original 

Analysis 

Original 

Analysis 

card 

card 

card 

card 

00 

OQ 

IS 

i 

3  • 

g  .    8 

s.aag 

•1.2 

S.aa| 

"5 
1 

id 

i*si 

5d 

I*j  | 

P 

"S  fl  2  i 

fa 

S'S"| 

fa 

! 

2.85 

0.40 

5.70 

0.30 

11 

0.40 

4.4 

0.80 

3.31 

2 

2.82 

0.90 

5.62 

0.68 

12 

0.40 

3.4 

0.80 

2.56 

3 

2.78 

1.23 

5.56 

0.93 

13 

0.40 

2.4 

0.80 

1.81 

4 

2.64 

1.40 

5.28 

1.05 

14 

0.40 

1.2 

0.80 

0.93 

5 

2.01 

1.90 

4.02 

1.43 

15 

0.51 

0.9 

1.02 

0.68 

6 

1.56 

2.40 

3.12 

1.81 

16 

0.80 

0.6 

1.60 

0.45 

7 

1.28 

2.90 

2.56 

2.18 

17 

1.18 

0.4 

2.36 

0.31 

8 

1.10 

3.40 

2.20 

2.56 

18 

1.50 

0.4 

3.00 

0.31 

9 

0.93 

4.00 

1.86 

3.00 

19 

2.00 

0.4 

4.00 

0.31 

10 

0.70 

4.20 

1.40 

3.16 

20 

2.50 

0.4 

5.00 

0.31 

FIG.  82. — Construction  of  T-S  diagram  from  p-v  diagram. 

This  method  considers  only  1  Ib.  of  steam  to  be  present  on 
the  expansion  line. 

The  p-v  diagram  is  now  laid  out  from  the  table  giving  the  fig- 
ure shown.  This  is  then  transferred  from  the  p-v  to  the  T-s 
quadrant  by  drawing  a  constant  pressure  line  until  it  strikes  the 


192  HEAT  ENGINEERING 

saturation  line  in  the  T-p  quadrant  and  this  fixes  the  tempera- 
ture of  the  point  A.  A  vertical  line  from  this  temperature  fixes 
a  line  in  the  T-s  quadrant  on  which  the  point  will  lie.  It  would 
be  at  a  if  it  had  a  quality  of  zero  and  at  6  if  of  quality  1,  and  in 
the  p-v  quadrant  it  would  be  at  c  if  of  zero  quality  and  at  d  if 
of  unit  quality. 

Now  v  -  v'  =  xv"  (41) 

and  s  -  s'  =  Y  (42) 

In  other  words  the  volume  change  and  the  entropy  change  from 
liquid  to  steam  depends  on  x.  Hence  the  point  A'  in  the  T-s 
quadrant  must  divide  a-b  in  the  same  way  that  A  in  the  p-v 
quadrant  bisects  c-d.  To  construct  such  a  point  the  point  of 
intersection  e  of  a  horizontal  from  b  and  a  vertical  from  d  is 
joined  to  the  intersection  /  of  a  horizontal  from  a  and  a  vertical 
from  c.  The  line  ef  is  then  the  quality  transfer  line  for  if  a 
vertical  is  drawn  from  A  to  this  line  intersecting  it  in  h  and  then 
a  horizontal  is  drawn  from  h  to  ab,  this  will  fix  the  point  A'. 
The  same  pressure  line  and  construction  transfers  B  to  B'.  If 
this  is  carried  on  in  the  same  manner  for  all  points  a  figure  such 
as  shown  in  the  T-s  quadrant,  or  enlarged  in  Fig.  83,  is  found  in 
the  T-s  quadrant  corresponding  to  the  p-v  diagram.  The  scales 
to  which  the  p-v  quadrant  has  been  drawn  were 

1  in.  =  2.5  cu.  ft. 

1  in.  =  20  Ibs.  per  sq.  in. 

The  T-s  quadrant  was  drawn  with 

1  in.  =  0.25  units  of  entropy 
1  in.  =  25°  F. 

The  area  scales  are: 

2.5  X  20  X  144 

1  sq.  in.  =  —        __Q        —  =  9.25  B.t.u.  (p-v) 

/7o 

1  sq.  in.  =  25  X  0.25  =  6.25  B.t.u.  (T-s) 

The  area  of  the  indicator  card  on  the  p-v  plane  was  7.94  sq. 
in.  and  on  the  T-s  plane  the  area  was  11.76  sq.  in. 

These  each  reduce  to  73.5  B.t.u.  for  the  area. 

The  T-s  plane  for  Fig.  83  has  been  turned  through  90°. 
This  is  the  conventional  T-s  diagram  for  the  indicator  card. 


THE  STEAM  ENGINE 


193 


It  is  now  necessary  to  add  other  lines  to  the  T-s  diagram  in 
order  to  interpret  it.  Letters  have  been  added  to  the  diagram 
at  critical  points  of  Fig.  83.  The  diagram  is  for  1  Ib.  of  total 
steam  and  this  is  the  amount  present  on  be  only  hence  the  figure  is 
true  only  for  this  short  line.  From  the  point  of  admission  /  to 
b  there  is  a  changing  weight  of  steam  and  from  c  to  e  there  is 
a  changing  weight.  The  points  on  these  lines  are  only  drawn 
according  to  the  actual  volumes  which  are  proportional  to  en- 
tropies because  each  is  proportional  to  MX.  To  study  the  line 


ef  on  which 


. 

i~ 


pounds  are  present,  the  line  corresponding 


to  zero  entropy  change  must  be  drawn.     If  AC  be  drawn  so 
that 

EC  M  0.0816         9Q 


E  B 


0.0905 


200  F, 


FIG.  83. — T-S  diagram  of  indicator  card. 

then  line  AC  would  be  the  liquid  line  for  the  boiler  steam  only, 
while  the  distance  CB  or  the  distance  between  AC  and  A  'B  repre- 
sents entropy  of  the  liquid  for  the  clearance  steam. 

The  distance  eF  represents  the  entropy  of  the  clearance  steam 
at  the  beginning  of  compression  and  GH  is  laid  off  at  this  distance 
from  CFA.  This  line  would  be  the  line  of  compression  if  the 

13 


194  HEAT  ENGINEERING 

clearance  steam  did  not  change  in  entropy  because  the  entropy 
of  the  liquid  CB  for  this  steam  plus  the  entropy  of  vaporization 
BH  is  constant.  GH,  although  not  an  adiabatic  on  this  figure, 
is  the  line  of  compression  if  the  clearance  steam  were  com- 
pressed without  giving  up  heat.  The  line  of  compression  is 
seen  to  pass  to  the  left  of  the  line  GH  which  means  that  heat  is 
taken  up  by  the  cylinder  walls.  The  amount  of  heat  is  found 
by  drawing  fg  parallel  to  GH  to  the  back  pressure  line  ed  and 
then  dropping  the  perpendiculars  to  h  and  i  at  absolute  zero  of 
temperature  getting  the  area  hgfei  as  the  heat  removed  in  com- 
pression. To  find  the  heat  lost  during  admission  from  /  to  b 
it  is  necessary  to  find  the  position  which  should  have  been  occu- 
pied by  the  steam  entering  from  the  boiler  if  no  heat  had  been 
taken  from  the  clearance  steam  or  the  steam  entering  the  en- 
gine. It  is  necessary  to  mark  the  line  corresponding  to  the  pres- 
sure at  the  throttle  valve  and  draw  this  on  the  diagram.  This 
will  be  the  line  ECBH.  On  this  line  EC  is  the  entropy  of  the 
liquid  of  the  boiler  steam,  CH  is  the  entropy  of  the  clearance 
steam  and  in  addition  to  this  there  must  be  HJ  for  the  vaporiza- 
tion of  the  steam  from  the  boiler.  If  x  is  the  quality  of  this 
steam  the  distance  HJ  is  equal  to 

(44) 


~  M  +  Mt 

T 

Since  BI  is  the  value  of  ™  for  1  Ib.  of  total  steam, 
BI  =  BH  +  #7 

The  point  which  should  have  been  at  /  is  now  at  b  and  the  area 
nJlk  is  called  the  loss  due  to  initial  condensation.  Since  the 
reference  line  to  care  for  other  losses  is  shifted  to  gM  the  point 
J  is  moved  to  K  and  the  loss  due  to  initial  condensation  is 
nKtk.  The  area  mnba  is  called  the  loss  due  to  wiredraw- 
ing. There  is  still  a  further  loss  fMmaf  which  occurs  during  the 
early  part  of  admission. 

The  area  prbk  is  the  loss  to  the  cylinder  walls  after  cut-off 
and  area  prc'o'  represents  a  gain  from  the  cylinder  walls,  and 
occ'o'  represents  a  loss  during  the  last  part  of  expansion. 

ksc'co  —  brs  =  net  gain  from  cylinder  walls  during  expansion. 

iedco  =  heat  removed  during  exhaust  and  dcu  =  loss  from 
incomplete  expansion. 


THE  STEAM  ENGINE 


195 


Now  the  line  GH  is  a  liquid  line  for  the  cylinder  feed  and 
hence  ieHJl  =  the  heat  supplied  with  the  cylinder  feed  for  one 
pound  of  total  steam,  above  the  temperature  of  the  exhaust. 

The  work  is  efabcd 

Hence 

ieHJl  +  (ksc'co  —  srb)  —  dbcdef  —  iedco  —  hgfei  =  heat  removed 
during  admission. 

This  will  be  found  to  be  equal  to  the  area  of  fMKtkbaf  if  gf 
is  continued  to  M  and  JK  is  made  equal  to  ME. 

The  various  losses  are  given  as  follows  (numerical  values 
should  check  with  Hirn's  analysis) : 


Area 

Sq.  in. 

Per  cent, 
of  work 
on  T-s 

B.t.u. 

Heat  supplied  above  exhaust  =  ieHJl  or  hgMKt  .  . 
Work  =  efabcd               .      

144.98 
11.76 
0.71 
0.30 

44.32 

2.80 
1.64 
8.43 
4.74 

1232.00 
100.00 
6.04 
2.55 
376.20 

23.80 
13.95 
717.50 
40.40 

905.00 
73.50 
4.45 
1.88 
277.00 

17.50 
10.30 
527  .  00 
29.60 

Loss  during  first  admission  =  fMma 

Loss  during  throttling  =  amnb  

Loss  due  to  condensation  =  nKtk 

Heat    added    from    walls    during    expansion  = 
ksc'co  —  brs                       

Loss  due  to  free  expansion  =  dcu 

Heat  necessarily  removed  during  exhaust  =  ieuo  .  . 
Heat  removed  during  compression  =  hgfi 

The  quality  of  the  steam  at  cut-off  is  given  on  Fig.  83  by  the 
ratio 


vb       28.8          n  r 
x  =  -  -  =  TT-X  =  0.645 
wv       44.6 


(45) 


and  this  can  be  found  in  the  same  manner  for  any  point  of  the 
expansion  curve.  For  the  compression  curve  x  may  be  found  by 
drawing  from  e  a  line  eO  so  that  its  distance  from  the  line  AC 

is  M   i  °T/T  times  the  distance  from  A'B  to  IN.     Thus 

M  +  Mo 


BO  =  BI 


Mt 


If  the  ratio  of  the  distance  from  AC  to  the  compression  line 
to  the  distance  from  AC  to  eO  at  the  same  level  be  found  this 
ratio  is  x.  AC,  A'B  and  eO  make  an  entropy  diagram  for  the 
compression  steam  with  a  bent  axis  AC. 

As  before  mentioned  the  only  true  lines  from  which  x  can 


196  HEAT  ENGINEERING 

be  found  are  be  and  ef.  The  other  lines  are  obtained  by  following 
a  scheme  in  which  the  product  MX  at  any  point  could  be  found 
but  not  either  of  the  terms  of  the  product.  Hence  one  cannot 
tell  what  losses  occur  at  various  points  on  this  line.  There  should 
be  an  agreement  by  the  two  analyses. 

MISSING  QUANTITY  AND  INITIAL  CONDENSATION 

Steam  of  quality  0.99  was  admitted  into  a  cylinder  and  after 
entering  the  cylinder  it  was  found  that  the  quality  at  cut-off  has 
been  reduced  to  0.633.  In  other  words  37  per  cent,  of  the  mix- 
ture is  moisture.  This  has  been  caused  by  the  action  of  the 
cylinder  walls.  Fig.  84  shows  the  actual  form  taken  by  the 
indicator  card  of  an  engine.  The  events  in  most  cases  take 

place  slowly  giving  rounded 
corners;  there  is  a  drop  due 
to  throttling  on  the  steam 
line  and  release  occurs  before 
the  end  of  the  stroke. 

If  the  weight  of  dry  steam 
shown  by  the  cards,  Fig.  84, 
is  found  by  the  formulae  be- 
low and  this  is  subtracted 

FIG.  84.-Actual    indicator    card   from    frOm    *he   am°Unt   °f   steam 
engine.  actually  supplied,  the  differ- 

ence is  called  the  "missing 

quantity."  It  is  expressed  as  a  percentage  of  the  steam  actually 
supplied.  It  is  caused  by  initial  condensation  of  the  steam. 

^  =     °'  x     -  ° 


Ma  —  apparent  weight  from  diagram. 

F  =  area  of  piston  in  square  feet. 

L  =  length  of  stroke  in  feet. 

D  =  length  of  card  in  inches. 
I  =  per  cent,  clearance  from  Fig.  84. 

v"  =  specific  volume  of  dry  steam. 
i,  «2  =  distances  from  clearance  lines  to  points  in  inches. 

If  M  equals  the  real  weight  taken  per  card, 

M-Ma  Mm 

~M  —         ~~M~  =  m"(i">  missmg  quantity. 


THE  STEAM  ENGINE  197 

This  is  almost  equal  to  1  —  #2  from  either  of  the  previous 
analyses. 

This  initial  condensation  is  due  to  the  effect  of  the  cylinder 
walls.  As  the  steam  in  the  cylinder  drops  in  pressure  its  tem- 
perature changes  and  the  moisture  on  the  walls  of  the  cylinder 
from  condensation  is  evaporated  removing  heat  from  the  metal 
and  moisture  and  cooling  them.  During  the  exhaust  this  action 
continues,  the  walls  being  so  cooled  that  when  fresh  steam  enters 
the  cylinder  some  of  the  steam  is  condensed  and  settles  on  the 
walls  of  the  cylinder.  This  enables  the  walls  to  absorb  heat 
from  the  steam  at  a  faster  rate  than  it  would  were  dry  vapor  in 
contact  with  the  metal.  This  condensation  produces  the  miss- 
ing quantity  and  the  presence  of  a  film  on  the  walls  of  the  cylinder 
seems  to  make  it  easier  to  carry  heat  to  the  walls  during  admis- 
sion while  during  the  exhaust  the  possibility  of  evaporating  this 
water  means  that  during  this  time  much  heat  is  given  up  by  the 
walls  and  discharged  with  the  exhaust  steam  when  it  can  be  of 
no  value  for  driving  the  engine. 

If  superheated  steam  be  furnished,  it  will  give  up  its  heat  to 
the  cylinder  walls  but  the  absence  of  the  moisture  film  makes 
this  action  slower  and  if  the  superheat  is  sufficient  the  necessary 
heat  may  be  abstracted  before  the  steam  is  reduced  to  the 
saturated  condition  or  before  much  condensation  occurs.  At 
release  the  absence  of  considerable  moisture  in  the  cylinder 
prevents  the  removal  of  much  heat  during  the  exhaust.  Thus 
the  superheated  steam  prevents  the  excessive  abstraction  of 
useful  heat  by  the  metal  during  admission  and  its  restoration 
when  it  cannot  be  used.  It  is  for  this  reason  that  superheated 
steam  is  such  a  valuable  medium  to  use.  This  accomplishes  a 
greater  increase  in  efficiency  than  that  which  is  shown  by  theo- 
retical considerations. 

To  determine  the  amount  of  steam  used  by  an  engine  it 
will  be  necessary  to  know  this  missing  quantity.  Then 

Ma 

M  =  z 

1  —  m.q. 

To  find  the  value  of  m.q.  theoretical  and  empirical  formulae 
have  been  proposed. 

It  will  be  evident  that  the  amount  of  condensation  per  stroke 
must  depend  on  the  surface  exposed  to  the  steam  at  cut-off, 
the  temperature  range  and  the  time  during  which  this  surface 


198  HEAT  ENGINEERING 

is  exposed  to  the  steam  or  inversely  on  the  number  of  times 
it  is  exposed  to  the  action  of  the  steam  per  minute.  There 
have  been  several  formulae  proposed.  Some  are  quite  complex 
taking  into  account  many  variables.  The  more  complex,  al- 
though they  may  be  correct  in  theory,  are  so  changed  by  slight 
changes  in  conditions  that  the  possibility  of  error  is  as  great  as 
in  the  less  complicated  formulae.  Some  of  the  formulae  pro- 
posed are  as  follows : 

m.q.  217 (r  -  0.7) 


, 

(Perry)  (47) 

m.q.  15(1  +  r 

- 


100  log  r 

~~  (Cotterill)  (48) 


(Thurston)  (49) 


1  -  m.q. 

m-1-  ~  inO-4-7    (Callendar  and  Nicolson)       (50) 


100  m.q.  =  1  '-  (Rice)  (51) 


Pressure                                 a                               /8  a 

60                                  0.568                         0.412  133 

80                                  0.517                         0.384  106 

100                                  0.466                         0.359  87 

120                                 0.414                         0.333  78 

140                                 0.363                         0.306  75 

In  the  above  formulae, 

r  =  ratio  of   expansion,  the  volume  at  end   of 
stroke  divided  by  the  volume  at  cut-off. 

d  =  diameter  in  inches. 

pi  =  absolute  steam  pressure  in  Ibs.  per  sq.  in. 
N  =  r.p.m. 

I  =  percentage  clearance. 


m.q.  =  c[Tb  -  Teldcos-1—-  +  [Tb  -  Te] 

* 

(127c  +  0.055)s         (Marks)  (52) 


c  =  0.02. 

b  =  abs.  temp,  at  cut-off. 


THE  STEAM  ENGINE 


199 


Te  =  abs.  temp,  during  exhaust. 
d  =  diam.  in  feet. 
e  =  fraction  of  volume  at  cut-off. 
s  =  stroke  in  feet. 
b  =  fraction  of  volume  at  compression. 

7 

/TT        i    \  /r*o\ 

i    r-l  no  IT"  I  I   ^  X  I 

~"   I  XAv?L/lvy  \OOy 

m.q.  =  missing  quantity  for  small  compression  or 

(1  —  x)  at  cut-off  for  large  compression. 
N  =  r.p.m. 
s  =  nominal  cylinder  surface  divided  by  volume, 

each  in  feet, 


'>  -\L  +  d)- 


T  =  temperature  range  from  limiting  pressures 

on  curve  Fig.  85.     (Ti-T*) 
p  =  abs.  pressure  at  cut-off  in  Ibs.  per  sq.  in. 
e  =  ratio  of  total  volume  at  cut-off  to  volume 

swept  out  by  piston. 


Absolute  Pressure  of  Top  and  Bottom  Lines  of  Indicator 
Cards  in  Pounds 

o  8  g  ^  i  £  g  £ 

/ 

1 

I 

/ 

/ 

/ 

/ 

/ 

/ 

50    100    150    200    250    300    350    400   45C 

Values  of  T 

FIG.  85. — Heck's  value  of  T  for  different  pressure. 

The  basis  for  the  form  of  these  formulae  may  be  given  as 
follows : 

The  surface  exposed  to  the  steam  at  cut-off  in  square  feet  is 


d  =  diam.  in  feet. 


200  HEAT  ENGINEERING 

L  =  stroke  in  feet. 
-  =  point  of  cut-off. 
Fp  =  surface  of  passages. 
The  time  during  which  this  is  exposed  is 

k  M.  minutes 
N 

where  k  =  a  fraction  less  than  1 

N  =  r.p.m. 

The  temperature  range  for  this  surface  is  some  function  of 
T.  It  is  not  equal  to  the  actual  difference  in  the  temperatures 
at  cut-off  and  compression  as  the  range  in  temperature  in  the 
metal  is  much  less  than  this.  The  quantity  of  steam  necessary 
to  give  the  heat  per  stroke  required  by  the  cylinder  walls  will 
be  proportional  to  the  surface,  time  and  temperature  range,  or 

M.  =  VF,f(T)  (55) 


To  express  this  as  a  fraction,  m.q.,  of  the  steam  shown  by  the 
indicator  card  it  must  be  divided  by  the  steam  shown  by  the  card. 
This  steam  is  given  by 


7*  L  4 
Ma  =  -     —r.  ---  compression  steam.  (56) 

V    2 

k" 
Now    the    specific    volume  v"2  =  —  approximately    since    the 

curve  of  saturated  steam  on  the  p-v  plane  is  nearly  a  rectangular 
hyperbola.     (See  Fig.  96.) 


Ma  =  /b'" 

T        TD 


Hence 

M8 


Now     ,2C     may  be  called  s.     Fc  is  practically  equal  to  2  -  —  f- 


*-i  T! 

irdL,  so  that  '  s  =  Y  +  ^ 

L       a 


THE  STEAM  ENGINE  201 

This  is  a  function  of  -3  giving 

'  m.q.  =  fc"  ±  5j/(D  or  ff  ±  p/(D  (58) 

If  r  is  written  as  -,  e,  being  the  relative  total  volume  at  cut-off 

6 

this  reduces  to 


By  examining  a  number  of  tests  of  all  sorts  of  engines  and  with 
different  speeds,  pressures  and  cut-offs  for  the  same  engine, 
investigators  have  been  led  to  the  empirical  forms  shown  above. 
What  in  the  simple  theory  appears  as  the  first  power  has  in 
general  been  changed  to  a  fractional  power.  In  all  of  the  simpler 
formulae  the  form  agrees  in  the  positions  of  the  terms  with  the 

k 
simple  theoretical  form  shown.     In  place  of  -^  being  used  for  the 

time  term  —7=  or  -77=  has  been  used.     Heck  assumes  that  the 

VN 


last  two  terms  of  Fc  are  equal  to  irdl,  thus  making  Fc  equal  to  the 
inside  surface  of  the  piston  displacement, 


From  evidence  shown  by  those  who  have  derived  these  for- 
mulae it  appears  that  the  formula  of  Heck 

0.27      Mr 

m-q- ="  w  V^  (57) 

is  a  good  one  to  use.  It  is  applicable  to  non-jacketed  engines 
when  supplied  with  dry  steam.  For  this  reason  it  cannot  be 
used  when  the  supply  is  superheated  steam  nor  for  the  lower 
cylinders  of  multiple  expansion  engines  when  the  steam  supply 
is  quite  wet.  If,  however,  this  steam  is  dried  by  a  separator  or 
by  reheater  coils  before  entering  the  other  cylinder,  it  may  be 
used.  For  jacketed  engines  it  might  be  used  by  deducting  the 
surfaces  heated  by  the  jacket  from  the  area  Fc  in  finding  this  or 
the  quantity  s. 

This  formula  shows  that  the  proportional  amount  of  condensed 
steam  varies  inversely  as  the  square  root  of  the  cut-off  getting 
less  as  the  cut-off  becomes  greater.  This  must  not  be  confused 


202 


HEAT  ENGINEERING 


with  the  actual  amount  of  steam  condensed  which  will  not  vary 
very  much  as  the  cut-off  changes.  The  large  part  of  the  conden- 
sation takes  place  when  the  steam  first  enters  the  cylinder  and 
as  the  piston  moves  along  the  additional  amount  is  not  great. 
The  surfaces  which  are  most  important  in  the  condensation  of 
steam  are  the  cylinder  head  and  piston.  For  this  reason  greater 
gain  is  to  be  expected  from  the  jacketing  of  the  heads  than  from 
the  jacketing  of  the  barrel  of  the  cylinder.  It  might  pay  to 
supply  the  hollow  piston  with  steam,  making  a  jacket  of  the  cored 
spaces.  The  value  of  s  decreases  as  the  size  of  the  cylinder  in- 
creases giving  less  condensation  in  large  cylinders  than  that 
found  in  small  ones. 

This  is  to  be  applied  to  the  card  below,  Fig.  86,  which  is  the 
assumed  card  in  the  design  of  a  20-in.  X  24-in.  engine  to  run  at  80 
r.p.m.  The  clearance  is  7  per  cent.;  cut-off  is  at  25  per  cent. 
Pressure  at  cut-off  is  125  Ibs.  abs.  and  back  pressure  17  Ibs.  abs. 


20  =  1+2.4=3.4 
12 


=   24" 
12 

L25    =    360 

\7  =  217  t 
T  =  360  -  217  =  143 
e  =  0.25  +  0.07  =  0.32 
0.27 


From  curve,  Fig.  85. 


m'q'  -- 


/3.4 
\125 


X143 


X0.32 


=  0.219 


FIG.  86. — Card  for  proposed    20  X  40  engine. 
Amount  of  steam  indicated  at  cut-off  is  given  by 


=  (0.25  +  0.07)  |^X 
0.32  X  4.36 


3.581 


144 
0.390  Ibs. 


'l25 


THE  STEAM  ENGINE  203 

c\  QQ  N/  A  Q£i 
Steam  at  compression  =  --  00      '  -  =  0.0615 

Zo.oo 

Indicated  steam  per  hour  =  (0.390  -  0.0615)  X  80  X  2  X  60 
=  3160  Ibs. 

S160 

Probable  steam  per  hour  ^  —  pr^TFrk  =  4000  Ibs. 

i  —  u.^iy 

Mean  height  =  1.41  in. 
m.e.p.  =  56.4 


56.4  X        X  314  X  80 

3300 
Probable  steam  consumption, 


H'p-  =  33000  -  X  2  =  172 


4000 
Steam  cons.  =         -  =  23.2 


This  result  is  low  so  that  the  missing  quantity  will  be  com- 
puted by  some  of  the  other  formulae. 
By  Perry's  formula  (48) 


1  -  m.q.         20  X   V80 
m.q.  =  0.271 

By  Thurston's  formula 

30*  '^ 


--    =  0.37 


1  -  m.q.      20  X  A/80 
m.q.  =  0.237 

By  Cotterill's  formula 

100  log 


=  0.293 


= 
1  -  m.q.        20  X  V80 

m.q.  =  0.226 

The  result  of  23.2  Ibs.  per  hour  per  horse-power  was  low  for 
this  type  of  engine  so  that  Heck's  formula  would  not  give  as  good 
results  as  Thurston's,  Perry's  and  Cotterill's.  If  0.30  is  taken 
for  the  value  of  mq.  The  steam  consumption  is  equal  to 


204  HEAT  ENGINEERING 

On  account  of  valve  leakage  this  result  would  be  increased 
slightly. 

EXPERIMENTS  OF  EFFECT  OF  CYLINDER  WALLS 

This  influence  of  the  cylinder  walls  has  been  experimentally 
studied  by  a  number  of  investigators.  Callendar  and  Nicolson, 
in  their  paper  presented  in  the  "  Proceedings  of  the  Institution 
of  Civil  Engineers  of  Great  Britain,"  Vol.  cxxxi,  gave  results 
obtained  by  them  on  a  10^  X  12  Robb  engine  in  1895  with  a 
flat  slide  valve  behind  a  pressure  plate.  The  piston  displace- 
ment was  0.601  cu.  ft.  and  the  clearance  volume  was  0.060  cu. 
ft.  The  engine  was  made  single-acting  so  as  to  study  the  action 
in  a  better  manner.  To  find  the  temperature  of  the  metal  in 
the  cylinder  head  eight  holes  were  drilled  at  regular  intervals  at 
1^2  in.  from  the  center  of  the  head  but  extending  to  different 
depths.  The  thicknesses  of  metal  remaining  at  the  bottom  of 
seven  of  these  holes  were  0.01  in.,  0.02  in.,  0.04  in.,  0.08  in.,  0.16 
in.,  0.32  in.  and  0.64  in.  Along  the  length  of  the  cylinder  at  the 
end  of  stroke  and  at  4,  6  and  12  in.  from  it  a  pair  of  holes  were 
drilled  in  the  side,  one  to  within  0.04  in.  of  inner  surface  and 
the  other  %  m-  At  2,  8,  10,  14  and  16  in.  from  the  end  single 
holes  were  drilled  leaving  ^  m-  a^  bottom.  There  were  also 
four  holes  at  the  middle  of  the  stroke  distributed  around  the 
circumference  of  the  cylinder  extending  to  within  J/£  in.  of  the 
inner  surface  and  finally  three  vertical  holes  two  inches  deep  were 
drilled  along  the  side  of  the  barrel  at  1  in.,  7J/£  in.  and  15  in.  for 
the  use  of  mercurial  thermometers  or  platinum  resistance  ther- 
mometers, while  in  the  other  holes  thermocouples  were  used. 

The  thermocouples  were  made  by  soldering  wrought-iron 
wires  at  the  bottom  of  the  holes.  To  make  the  cold  junction, 
the  same  kind  of  wire  was  attached  to  cast-iron  blocks  cast  from 
the  same  ladle  as  that  from  which  the  head  was  cast.  These 
were  immersed  in  a  paraffine  bath  at  212°  F.  The  potential 
was  measured  by  a  very  sensitive  galvanometer  by  balancing 
the  potential  on  a  potential  wire.  The  formula  for  the  voltage 
in  microvolts  produced  by  this  couple  was  found  to  be  given  by 

E  =  1692  -  17.86*  +  0.0094Z2  (58) 

if  the  cold  junction  was  at  100°  C.  and  the  hot  junction  at  Z°C. 
Later  one  of  the  couples  in  the  cylinder  was  used  as  the  cold 


THE  STEAM  ENGINE  205 

junction  and  the  drop  of  temperature  between  these  two  was 
measured. 

This  method  was  somewhat  similar  to  that  used  by  Prof. 
E.  Hall  of  Harvard  (Trans.  A.  I.  E.  E.,  1891)  but  his  results 
were  not  very  extensive. 

To  find  the  temperature  of  steam  the  authors  used  a  platinum 
resistance  thermometer  in  the  cylinder  3  in.  from  the  face  of  the 
piston  and  also  one  in  a  small  %-in.  hole  in  the  center  of  the 
piston  head. 

To  get  the  temperature  at  a  definite  point  in  the  stroke  by 
any  couple  or  platinum  thermometer  a  pair  of  revolving  brushes 
were  attached  to  the  shaft.  One  brush  made  contact  with  a 
central  copper  tube  and  the  other  with  a  sector  mounted  on  a 
circular  disc.  The  sector  was  one-thirtieth  of  a  circumference 
in  length.  The  disc  could  be  rotated  and  by  a  scale  and  vernier 
the  position  of  the  crank  for  any  observation  could  be  read.  A 
number  of  sectors  on  the  disc  reduced  the  amount  of  motion  neces- 


36         39  40 


FIG.  87. — Indicator  card  marked  with  points  at  various  sixtieths  of  a  revolu- 
tion.    From  20  X  40  engine. 

sary.     To  facilitate  the  change  of  circuit  to  different  couples, 
mercury  cups  were  used. 

To  show  the  results  of  these  tests,  Fig.  88  has  been  constructed 
from  the  card,  Fig.  87  by  finding  the  saturation  temperature  for 
pressures  of  the  steam  at  points  corresponding  to  definite  crank 
angles.  This  temperature  from  the  steam  tables  is  plotted  to 
sixtieths  of  a  revolution,  giving  the  curve.  The  marks  X  give 
points  similar  to  the  results  shown  by  the  platinum  thermometer 
while  the  points  marked  0  show  the  temperatures  of  the  ther- 
mometer in  the  steam  space  in  the  cylinder  head.  The  curves 
illustrating  the  variation  of  temperature  of  the  metal  at  J£5  in. 
from  the  inside  surface  in  the  head  and  that  at  holes  in  the  side 
at  4  in.  from  end  are  shown  to  a  larger  scale  above  the  card. 
These  curves  are  ideal  and  are  drawn  to  indicate  the  results  of 
Callendar  and  Nicolson. 


206 


HEAT  ENGINEERING 


The  surprising  results  of  the  actual  tests  by  Callendar  and 
Nicolson  were  that  at  Jf  oo  in.  from  the  inside  surface  the  variation 
of  temperature  was  only  4.3°  F.  at  100  r.p.m.,  while  at  ^5  in. 
from  inside  the  variation  was  6°  F.  at  46  r.p.m.  and  4°  F.  at  73.4 
r.p.m.  At  J<2  in.  from  the  inside  the  change  in  temperature  due 
to  cyclic  variation  was  practically  zero. 


Area    cbd  +  ale  =  1.89 

Sq.  In. 
=  Temp.byPt.Ih.ermometei  Attached  to  Piston 


03      6     9     12    15    18    21    24    27    30    33    36    39    42    45    48    51    54    57    60 

Sixtieth  of  One  Revolution 

I'-loJj   Rev. 

Fid.  88. — Diagram  of  temperature  at  various  positions  of  piston  for  dif- 
ferent movement  of  crank.  Temperatures  taken  in  head,  cylinder  wall  and 
steam  space.  (After  Callendar  and  Nicolson.) 


Along  the  length  of  the  barrel  the  first  hole  situated  in  the 
clearance  space  and  exposed  to  the  same  steam  action  as  the 
heads  showed  13.5°  F.  variation  at  ^5  in.  from  inside  at  44  r.p.m. 
when  the  heads  at  this  thickness  showed  4.9°  F.  variation. 
This  amount  of  13.5°  F.  at  J^5  in.  from  the  surface  probably  corre- 
sponds to  about  20°  F.  variation  at  the  surface  and  this  particular 
result  was  the  largest  obtained  by  the  experimenters.  The 


THE  STEAM  ENGINE  207 

variation  at  4  in.  was  found  to  be  5°  F.  and  at  6  in.,  3^° F.  It  was 
found  that  these  temperatures  begin  to  rise  before  the  piston 
reaches  these  points  showing  that  there  was  conduction  along  the 
barrel  and  also  the  steam  could  leak  around  the  piston  as  far  as 
the  rings.  A  curious  result  near  the  middle  of  the  stroke  is  the 
temperature  that  is  shown  on  the  inner  surface  which  is  lower  than 
that  some  distance  farther  out.  This  shows  that  there  is  a  flow 
of  heat  from  the  metal  to  the  steam  at  this  point.  The  experi- 
ments seem  to  show  a  gradient  of  0.55°  F.  per  inch  in  the  head 
while  that  on  the  side  wall  of  the  barrel  was  a  variable  quantity 
in  an  axial  direction  being  greatest  at  the  center  of  the  stroke 
where  it  was  9.3°  F.  per  inch. 

Experiments  were  made  to  determine  the  conductivity  of 
cast  iron  giving  5.5  B.t.u.  per  square  feet  per  degree  per  hour 
for  1  in.  thickness  and  then  the  diffusivity,  which  is  the  ratio  of 
this  conductivity  to  the  thermal  capacity  of  the  same  amount  of 
metal,  was  computed. 

The  results  of  these  experiments  showed  that  the  effect  of 
moisture  in  the  steam  is  to  increase  the  condensation  while 
superheat  decreases  it.  The  percentage  amount  of  condensa- 
tion varies  inversely  as  the  ratio  of  expansion  making  the  actual 
amount  of  condensation  practically  constant.  The  effect  of 
initial  pressure  is  complex  so  that  an  increase  does  not  mean 
an  increase  in  condensation  and  the  same  may  be  said  of  the 
change  in  back  pressure.  Early  compression  means  a  shorter 
time  for  cooling  and  hence  decreases  the  condensation.  Jacket- 
ing reduces  the  surface  which  causes  condensation  and  for  that 
reason  reduces  the  amount  of  condensation. 

To  find  the  maximum  condensation  possible,  the  limiting 
amount  as  they  called  it,  the  average  height  of  the  temperature 
degree  diagram,  Fig.  88,  is  found  and  when  the  area  above  this 
line  is  expressed  in  degrees  and  sixtieths  of  a  cycle  it  will  give 
the  thermal  units  per  hour  per  square  foot  when  multiplied  by 
45.  The  surface  considered  for  20  per  cent,  cut-off  on  this  en- 
gine which  was  a  Corliss  engine  was  8.9  sq.  ft.  on  each  end. 
45  X  Area  (cbd  +  abc)  X  scale  X  surface  of  clearance  =  45  X 
1.89  X  10  X  50  X  8.9  X  2  =  Total  B.t.u.  of  condensation  per 
hour  =  758,000  (59) 

This  quantity  is  then  multiplied  by 

1  +  VN        VN 

or  (60) 


208  HEAT  ENGINEERING 

to  make  it  correct  for  the  speed  of  N  revolutions  per  minute  for 
cut-off  or  release  respectively;  while  multiplying  by  1.4  gives 
the  result  for  a  double-acting  cylinder. 

Thus,  from  the  Fig.  88,  the  following  results  for  a  20  in.  X 
42  in.  engine  at  80  r.p.m.  : 


758,000  X  -    '    y     X  1.4  =  884,000  B.t.u.  per  hr. 
3  "fr  \/80 

If  the  pressure  at  entrance  from  the  card  were  100  Ibs.  absolute 
for  which  r  =  887.6,  the  steam  condensed  per  hour  will  be 


This  engine  uses  4000  Ibs.  of  steam  per  hour  so  that  the 
missing  quantity  is  0.25. 

The  actual  condensation  may  be  much  less  than  this  and,  if 
the  mean  temperature  of  the  cylinder  wall  is  known,  the  area 
above  this  line  to  any  event  gives  the  actual  condensation  when 
handled  in  the  manner  shown.  The  net  area  to  a  point  beyond 
the  crossing  of  the  mean  line  gives  the  difference  between  the 
condensation  and  the  reevaporation.  The  multiplier  is  different 
for  cut-off  and  release. 

VALVE  LEAKAGE 

Callendar  and  Nicolson  found  that  in  their  engine  much  of 
the  apparent  missing  quantity  was  due  to  leakage  of  steam 
beneath  the  valve  into  the  exhaust  and  this  steam  never  entered 
the  cylinder.  The  valve  leakage  of  steam  was  found  to  depend 
on  the  periphery  of  the  valve,  the  lap  and  the  pressure  difference 
or 

W  =  KI(P>  -  **>  (61) 

s 

when  pi  and  p2  were  the  mean  pressures  on  the  two  sides  in 
pounds  per  square  inch,  s  was  the  lap  in  inches  and  I  was  the 
periphery  in  inches. 

The  value  of  this  leakage  term  was  undoubtedly  large  in 
the  experiments  under  review  but  in  many  engines  it  may  be  a 
small  quantity.  In  the  engine  used  K  was  equal  to  0.02. 

The  platinum  thermometer  projecting  into  the  steam  of  the 
cylinder  3  in.  from  the  piston  face  indicated  (Fig.  88)  that 


THE  STEAM  ENGINE 


209 


ICO  Ca 


120 


0  C. 


Piston  Travel 


the  steam  was  practically  saturated  except  during  compres- 
sion when  it  was  superheated,  while  that  in  the  %-in.  hole  in  the 
piston  head  was  highly  superheated  during  all  of  the  stroke 
except  at  cut-off.  This  was  undoubtedly  due  to  the  superheating 
effect  of  the  metal  when  the  pressure  was  low  causing  the  steam 
to  be  superheated  still  further  during  compression. 

George  Duchesne  in  the  Revue  de  Mechanique1  gives  results 
of  an  investigation  with  a  hyperthermom- 
eter  which  consisted  of  a  multiple  silver- 
platinum   couple    0.002  mm.  thick.     This 
couple  responded  rapidly  to  the  changes  of 
temperature.     He  finds  the  cycle  of  Fig.  89  |ioo 
for  the  wall,  steam  and  the  saturation  tern-  |  so 
perature  of  the  steam  in  his  engine  on  a  I  eo 
temperature-piston     travel     diagram.     In     40 
this  it  is  noted  that  the  wall  temperature     20 
is  usually  higher  than  the  steam  and  that 
the  steam  is  saturated  except  during  com- 

pression when  it  is  superheated,  becoming    ^Fl^'  89>    RJsults  of 

r  ,  °    Duchesne  on  tempera- 

saturated  at  the  point  where  it  reaches  the    ture  of  wall  and  steam. 

wall   temperature.     Of   course   this   metal 

cycle  is  true  for  one  point  in  the  cylinder  but  for  another  point 
the  cycle  would  probably  be  lower.  Duchesne  states  that  the 
compression  should  not  be  carried  higher  than  the  temperature 
of  the  metal  of  the  head  for,  if  this  is  done,  a  loop  is  found  on  the 
card,  Fig.  90.  That  this  loop  is  due  to  excessive  condensation 
was  proven  by  the  fact  that,  when  air  was  used  with  this  amount 
of  compression,  the  loop  was  not  shown, 
compression  being  carried  up  to  a  higher 
pressure.  The  loop  may  not  be  found  in 
high-speed  engines  when  the  piston  speed 
is  faster  than  the  speed  of  condensation. 

This  compression,  however,  does  not 
affect  the  steam  consumption  very  greatly 
as  has  been  shown  by  Prof.  John  Barr2 
in  1895  and  by  E.  Heinrich3  in  1913.  In  these  experiments  there 
was  a  slight  change  in  the  steam  per  indicated  horse-power  hour 

1  Revue  de  Mechanique,  1897,  pp.  925  and  1236.  Power,  June  28,  1910; 
May,  1911. 

2  Trans.  A.  S.  M.  E.,  1895,  p.  430. 

3  Zeit.  des  Verein  Deutscher  Ing.,  Vol.  58,  Jan.  3,  1914. 

14 


FIG.    90.  —  Duchesne 


210 


HEAT  ENGINEERING 


Pressure  orTemperature 
"  Iteam  vs  Stroke 


Temperature  of  Metal 
vs  Stroke 

FIG.  91. — Adams 
results. 


with  change  in  compression  but  this  was  very  slight.  Heinrich 
found  that  by  adding  plates  and  increasing  the  clearance  surface, 
although  the  volume  remained  the  same,  the  water  rate  was 
increased,  showing  the  harmful  effect  of  sur- 
face. 

In   1895,  at  Cornell  University,  Mr.  E.  T. 
Adams  used  a  couple  in  the  cylinder  wall  near 
the  inside  surface  (Koo  in.)  and  attached  it  to 
a  galvanometer  of   short  period.     He   photo- 
graphed the  beam  of  light  from  the  mirror  of  the 
galvanometer  on  a  sensitive  paper  moved  by 
the  piston.     This  result  is  shown  in  Fig.  91, 
which  shows  a  rapid  drop  at  the  opening  of  re- 
lease probably  due  to  the  evaporation  of  water  from  the  surface 
and,  as  the  heat  flowed  in  from  the  outer  metal  of  the  wall, 
this  temperature  rose  again. 

STEAM  CONSUMPTION  BY  CLAYTON'S  METHOD 

J.  Paul  Clayton1  has  studied  the  results  of  actual  tests  of  engines 
and  has  shown  that  the  expansion  lines  of  steam  engines  and,  in 
fact,  the  expansion  lines  from  any  engine  using  an  elastic  medium 
are  polytropic  curves  of  the  form  pvn  =  K  unless  there  is  some 
peculiarity  such  as  leakage  or  some  other  disturbance  which  is 
not  uniform.  He  then  investigated 
the  values  of  n  from  different  en- 
gines, finding  the  value  of  n  by  plot- 
ting a  logarithmic  diagram  of  pres- 
sure and  volume  as  shown  for  the 
cards  of  Fig.  92  taken  from  a  20 
in.  X  42  in.  engine  at  78  r.p.m.  with 
4  per  cent,  clearance.  The  table 
below  shows  the  method  of  con- 
structing the  diagram  Fig.  93.  If  the  clearance  is  properly 
measured  and  the  correct  scale  of  the  spring  is  used,  the  ex- 
pansion lines  become  straight  lines  on  the  logarithmic  diagram 
and  the  slopes  of  these  lines  are  the  values  of  n  since 

log  pi  -  log  pz 
log  v2  -  log  vi 

1  Trans.  A.  S.  M.  E.,  34,  p.  17. 

Bulletin  58,  Engineering  Experiment  Station,  Univ.  of  Illinois. 
Bulletin  65. 


Spring 


FIG.  92. — Card  from  engine  for 
Clayton  analysis. 


n  = 


(62) 


THE  STEAM  ENGINE 


211 


The  variation  of  this  line  from  a  straight  line,  as  shown  in  the 
figure,  would  indicate  leakage. 


Point 

Volume  in  fifths 
inches  from 
clearance  line 

Logarithm  of 
volume  in    fifths 
inches  +  1 

Pressures  in  fifths 
inches  from 
absolute  zero 

Logarithm  of 
pressure  in  fifths 
inches  +  1 

1 

0.70 

0.846 

8.65 

.936 

2 

2.20 

1.340 

8.40 

.924 

3 

3.88 

1.590 

8.30 

.919 

4 

5.50 

1.740 

8.28 

.918 

5 

6.25 

1.795 

7.97 

.902 

6 

7.47 

1.873 

6.50 

.813 

7 

9.35 

1.970 

5.20 

.716 

8 

12.80 

2.106 

3.80 

.579 

9 

15.85 

2.200 

3.15 

.498 

10 

18.92 

2.276 

2.05 

.424 

11 

17.80 

2.250 

2.18 

.338 

12 

15.20 

2.182 

1.88 

.274 

13 

12.55 

2.099 

1.85 

.267 

14 

9.50 

1.978 

1.80 

.255 

15 

5.70 

1.756 

1.80 

.255 

16 

3.52 

1.546 

1.75 

.243 

17 

2.80 

1.447 

2.00 

.301 

18 

1.75 

1.240 

3.15 

1.498 

19 

1.25 

1.096 

4.35 

1.638 

20 

0.70 

0.846 

7.20 

1.864 

2.00 


£1.75 


1.25 


1.00 


0.75 


3  4 


1.25  1.50  1.75 

Logarithm  of  Volume 


2.00 


2.25 


2.50 


FIG.  93. — Logarithmic  diagram  of  steam  engine  indicator  card. 

In  preparing  the  table,  distances  to  the  true  zero  of  pressure 
and  of  volume  have  been  found  and  then  the  distances  from  these 
lines  to  any  point  have  been  measured  in  fifths  of  an  inch.  The 
logarithms  of  the  numbers  have  been  increased  by  unity  to  have 


212 


HEAT  ENGINEERING 


all  of  the  numbers  positive  for  plotting.  This  only  shifts  the 
lines  but  does  not  change  the  slopes. 

The  logarithms  are  then  plotted  giving  the  figure  which  also 
shows  the  true  point  of  closing  or  opening  of  the  valve  as  the  point 
at  which  the  straight  line  begins  marks  the  beginning  or  ending 
of  the  expansion  or  compression.  This  is  of  value  as  the  events, 
especially  compression  are  difficult  to  fix  accurately  on  the  indi- 
cator card. 

On  investigating  tests  on  which  the  quality  of  the  steam  in  the 
cylinder  could  be  determined  from  data  and  on  making  such 
tests  on  an  engine  in  the  laboratory  where  the  quality  of  steam 


0.85 


0.95 


1.15 


1.25 


50 


70 


90  110 

Initial  Steam  Pressure 


130 


150 


FIG.  94. — Clayton's  relations  between  n,  x,  and  p  for  stationary  engines. 

at  cut-off  could  be  varied  by  using  wet  steam  or  superheated 
steam,  Clayton  showed  that  there  was  a  variation  of  the  value 
of  n  of  the  expansion  line  and  the  quality  of  steam  at  cut-off. 
The  value  of  n  varied  from  0.70  to  1.34  while  the  quality  at  cut- 
off varied  from  about  0.4  to  0.9.  He  showed  that  there  was 
a  relation  between  them.  In  steam  engines  the  value  of  n  of  the 
expansion  line  could  be  told  from  the  quality  at  cut-off  or,  con- 
versely, knowing  n  the  quality  at  cut-off  could  be  found.  This 
relation  changes  with  the  steam  pressure  and  with  the  speed  but 
does  not  vary  with  the  point  of  cut-off.  Since  in  stationary  en- 
gines the  variation  with  the  speeds  in  use  is  slight,  the  only  varia- 
tion considered  is  the  variation  with  pressure  and  Fig.  94  has  been 
prepared  from  a  similar  figure  due  to  Clayton  giving  the  values 


THE  STEAM  ENGINE 


213 


of  x  and  n  at  different  pressures.  For  locomotive  work,  where  the 
pressures  do  not  vary  much,  Clayton  gives  a  figure  similar  to 
Fig.  95  showing  the  values  of  x  and  n  at  different  speeds. 

The  results  have  been  obtained  from  non- jacketed  engines 
exhausting  near  the  atmospheric  pressure  and  hence  the  results 
should  only  be  applied  to  such.  The  method,  if  used  with 
jackets  or  with  high  back  pressure  can  only  be  considered  an 
approximation.  The  indicator  must  be  connected  directly  to 
the  cylinder  without  the  use  of  piping  and  a  correct  reducing 
motion  with  no  stretch  nor  lost  motion  must  be  employed. 


0,85 


1.25 


100  150  200  250  300 

R.P.M. 

FIG.  95. — Clayton's  relations  between  N,  n,  and  x  for  locomotive  engines. 


On  actually  applying  this  method,  a  result  close  to  the  correct 
steam  consumption  may  be  found  by  assuming  dry  steam  at 
compression.  The  error  may  be  4  per  cent.  The  results  for 
Fig.  92  will  be  computed. 

Value  of  n  for  expansion  curve,  Fig.  93,  =  0.98 

Steam  pressure,  108  Ibs.  abs. 

x,  from  Fig.  94,  for  n  =  0.98  and  p  =  108,  is  0.65 

Pressure  at  cut-off,  93  Ibs.  abs. 

Length  to  cut-off,  1.22  in.;  to  comp.,  0.73  in.;  of  card,  3.14  in. 


Total  volume  at  cut-off  =  TTT,  X 


2.96  cu.ft. 


3.14  '       1728 
Weight  of  dry  steam  at  93  Ibs.  =  0.2112  Ibs.  per  cu.  ft. 

w  .  ,  ^     ,   4  a      2.96X78X60X0.2112 

Weight  of  steam  per  hour  at  cut-off  =  - 


0.65 


=  4470  Ibs. 


214  HEAT  ENGINEERING 

0.73       314  X  42 
Volume  at  compression  =  ^rr  X  — 1f70Q      =1.76 


Pressure  at  compression  =  24  Ibs.  abs. 

Weight    of    clearance    steam    per    hour  =   1.76  X  78  X  60 

X  0.0591  =  486 

Steam  supplied  =  4470  -  486  =  3984  Ibs.  per  hour 
H.p.  =  160  h.p. 

Steam  per  i.h.p.-hour  =  "TTT  =  24.8  Ibs. 


VALUES  OF  N  FOR  EXPANSION  LINES 

Clayton  has  found  by  logarithmic  diagrams  that  the  value  of 
n  for  the  expansion  line  of  the  steam  engine  varied  from  0.835 
to  1.234  and  that  the  average  with  saturated  steam  in  the  steam 
pipe  was  0.947  while  with  superheated  steam  the  value  was  1.056 
and  the  average  of  all  tests  was  1.004.  The  value  of  n  =  1 
which  is  so  common  in  the  theoretical  discussion  of  indicator 
cards  of  steam  engines  is  not  far  from  the  truth.  The  ease  of 
construction  of  the  rectangular  hyperbola,  pv  =  constant,  and  the 
simple  results  it  leads  to  in  practice  are  good  reasons  for  its 
common  use.  Of  course  it  has  no  theoretical  basis  and  is  only 
used  for  the  reasons  given  above. 

In  steam  engines  actually  examined  by  Clayton  the  expansion 
line  had  values  of  n  near  1  in  all  cases  except  poppet  valve 
engines,  where  n  rose  to  1.3  with  highly  superheated  steam  and 
in  the  Stumpf  Straight  Flow  Engine  with  superheated  steam 
where  n  was  1.2.  The  compression  lines  were  mostly  near  n  =  I 
although  some  values  much  lower  were  found. 

In  gas  engines  n  for  expansion  varied  from  1.09  to  1.36  while  n 
for  compression  varied  from  1.09  to  1.43.  Gtildner,  according 
to  Clayton,  gives  values  of  n  from  1.30  to  1.38  for  compression 
and  1.35  to  1.50  for  expansion.  The  latter  is  due  to  hot  water. 
He  states  that  Burst  all  gives  1.288  as  the  average  for  expansion 
and  1.352  for  compression  in  gas  engines. 

For  compressed  air  locomotives  Clayton  found  n  =  1.35  while 
on  air  compressors  it  was  1.26. 

For  ammonia  compressors  the  cards  examined  gave  values 
averaging  n  =  1.20. 

On  gas  compressors  the  value  is  n  =  1.14.  This  is  very 
low. 


THE  STEAM  ENGINE  215 

While  discussing  this  value  of  n  it  is  well  to  remember  that, 
although  the  equation  of  the  adiabatic  of  steam  is 

/    ,   xr 
s  +  -i  =  const,  or 


,    ,   xr         £cpdt 
+  ~T  +   )    T    =  c° 


the  line  is  sometimes  put  in  the  approximate  form 

pvn  =  const. 

n  =  -JT,  according  to  Rankine  (63) 

y 

n  =  1.035  +  O.lz,  according  to  Zeuner  (64) 

n  =  1.059  +  0.000315?  +  (0.0706  +  0.000376p)z 

according  to  E.  H.  Stone  (65) 

p  =  pressure  at  point  where  quality  is  x 
x  =  quality  at  highest  point 

For  superheated  steam, 

p(v  +  0.088) 1-805  =  const.  (Goodenough) 

EXPANSION  LINES 

It  will  be  well  at  this  point  to  examine  the  difference  between 
various  lines  of  expansion  which  may  occur  in  the  steam  engine 
cylinder  to  see  what  error  might  be  made  in  assuming  one  rather 
than  another  in  the  theoretical  discussion  of  indicator  cards. 
The  discussion  and  curves  will  also  indicate  how  close  the  lines 
approach  each  other. 

Using  the  results  of  Clayton,  suppose  that  an  engine  with  cut- 
off at  Y±  stroke  and  with  10  per  cent,  clearance  has  steam  of  a 
quality  of  0.70  at  115  Ibs.  absolute  pressure  and  it  is  required 
to  draw  the  following  lines  through  this  point: 

(a)  Rectangular  hyperbola 
(6)  Adiabatic 

(c)  Isodynamic 

(d)  Constant  steam  weight,  x  =  const. 

(e)  pvn     =  const,  of  Clayton,  n  =  1.02 
(/)  pv1-2  =  const. 

(0)  pv°-8  =  const. 

From  Clayton's  diagram,  n  =  1.02. 


216 


HEAT  ENGINEERING 


The  diagram  for  1  Ib.  of  steam  will  be  computed  and 
tabulated  below  for  pressures  at  20-lb.  intervals  to  points  beyond 
the  actual  length  of  card.  The  computations  for  the  first  points 
are  given. 

Vi  =  Mxv"  =  1  X  0.70  X  3.876  =  2.71  cu.  ft. 
pi  =  115  Ibs. 
For  p2  =  95  Ibs.  abs., 

(a)  F2  =  2.71  X  ^jr  =  3.28  cu.  ft. 

_  0.4881  +  0.70  X  1.1026  -  0.4699 

1.1363 
=  0.695 

F2  =  0.695  X  4.644  =  3.23  cu.  ft. 
(6a)  Stone's  value  of  n, 
n  =  1.059  -  0.000315  X  115  +  (0.0706  +  0.000376 

X  115)0.70 
n  =  1.102 


2.71 


=  3.23  cu.  ft. 


x2  X  808.8  =  309.0  +  0.70  X  797.0  -  294.6 
z2  =  0.708 

F2  =  0.708  X  4.644  =  3.28  cu.  ft. 
(d)  V2  =  0.70  X  4.644  =  3.25  cu.  ft. 


(e)  F2  =  2.71  1      =  3.28  cu.  ft. 


(/)  y2  =  2.71  (i^p  =  3.17  cu.  ft. 
TABLE  OF  VOLUMES 


Pressure 

n  =  l 

Adiabatic 

Isody- 
namic 

x  —  const. 

n  =  1.02 

ra  =  1.2 

n  =  0.8 

Theory 

Stone 

115 

2.71 

2.71 

2.71 

2.71 

2.71 

2.71 

2.71 

2.71 

95 

3.28 

3.23 

3.23 

3.28 

3.25 

3.28 

3.17 

3.41 

75 

4.15 

4.00 

4.01 

4.15 

4.07 

4.13 

3.88 

4.62 

55 

5.67 

5.30 

5.31 

5.66 

5.45 

5.60 

5.02 

6.82 

35 

8.90 

8.00 

8.00 

8.82 

8.32 

8.70 

7.34 

11.95 

The  tables  show  the  close  agreement  of  the  line  pv1-102  with  the 
adiabatic  and  Fig.  96  shows  how  closely  the  rectangular  hyper- 


THE  STEAM  ENGINE 


217 


bola,  the  Clayton  line,  the  adiabatic,  the  isodynamic  and  the 
constant  steam  weight  curves  come  together.  The  lines  vary 
slightly.  If  the  card  abed  is  worked  out  for  the  various  lines 
it  is  found  that  although  the  mean  effective  pressure  for  the  lines 
pv0-8  and  pv1-2  vary  12  per  cent,  from  the  mean  of  the  two,  the 
variation  on  the  lines  pv1-02  and  pv1'1  which  are  the  extremes  of 
the  remaining  lines  vary  from  the  hyperbola  by  5  per  cent.,  or, 
5  per  cent,  is  the  maximum  variation  in  general.  The  value  1.2 


120 


100 


CO 

JB 

a  60 

S 

£ 

«   40 

p 

1 

i 


I  =1.102 
Adiabatic 


2.0 


4.0  6.0  8.0 

Volume  in  Cu.  Ft. 


10.0 


12.0 


FIG.  96. — Variation  in  various  expansion  curves.  Constant  steam  weight 
curve  lies  between  adiabatic  and  pv1-02  =  const.  Isodynamic  can  not  be 
separated  from  curve  pv  =  constant. 

is  found  on  engines  using  superheated  steam  and  the  variation 
in  the  m.e.p.  for  the  card  with  "this  line  is  14  per  cent,  from  that 
with  the  rectangular  hyperbola. 

USE  OF  RECTANGULAR  HYPERBOLA 

From  the  above  it  will  be  seen  that  in  general  the  rectangu- 
lar hyperbola  will  give  areas  within  2  per  cent,  of  the  actual 
cards  for  ordinary  engines  although  with  superheated  steam  or 
very  wet  steam  the  use  of  the  rectangular  hyperbola  for  the  ex- 


218 


HEAT  ENGINEERING 


pansion  line  may  result  in  an  error  of  less  than  15  per  cent,  in 
the  area  of  the  indicator  card.  Since  the  errors  due  to  round- 
ing of  the  corners  or  the  intersection  of  the  various  cards  of  a 
multiple  expansion  engine  may  lead  to  as  great  errors,  the 
simplicity  in  the  calculations  and  constructions  when  this  line  is 
assumed  makes  this  curve  of  value  in  the  preliminary  design  of 
an  engine.  When  the  probable  condition  of  the  steam  is  known 
the  exact  exponent  may  be  used. 

CONSTRUCTION  OF  EXPANSION  CURVES 

To  draw  the  rectangular  hyperbola  the  graphical  construction 
shown  in  Fig.  97  is  used.  Given  the  original  point  1,  a  hori- 
zontal and  a  vertical  line  are  drawn  through  1.  If,  from  a,  b 
and  c,  on  the  horizontal  line,  slanting  lines  are  drawn  to  the  origin 


FIG.  97. — Construction  of  rectangular  hyperbola. 

of  pressure  and  volume,  they  will  cut  the  verticals  in  a',  &'  and  c'. 
Verticals  from  a,  b  and  c  intersect  horizontals  from  a',  b'  and  c' 
in  a",  b"  and  c"  which  are  points  on  the  rectangular  hyperbola. 

The  student  may  see  that  this  curve  fulfills  the  equation  of  the 
hyperbola  since 

Pi  =  Pa 

P'a    =   P"a   • 


From  similar  triangle 


V    a    =   Vc 

!^_  _  PI 

V'       ~    V' 


Substituting  the  values  above 


or 


P 


"  „"      . 


THE  STEAM  ENGINE 


219 


The  graphical  construction  usually  employed  for  the  poly- 
tropic,  pvn  =  constant,  is  open  to  an  accumulative  error  and  for 
that  reason,  although  correct  geometrically  any  slight  error  of 
one  point  is  increased  in  the  later  points  and  hence  the  best  way 
to  construct  the  curve  is  to  assume  a  ratio  of  two  pressures,  say 

—  =  r,  and  to  use  this  to  compute  all  successive  points. 

p*  =  rpl 


In  this  way  successive  pressures  can  be  read  from  a  slide  rule 
without  moving  the  middle  scales  by  setting  the  zero  of  the 
C  scale  at  the  value  of  r  on  the  D  scale. 


P2  =  /VA 

PI    W 


Since 


The  successive  volumes  are  given  by 


r"v2 


The  p's  and  v's  being  known,  the  curve  may  be  plotted  quickly. 
Thus  suppose  the  line 

pj,i.4  _  constant 

is  desired  to  pass  through  p  =  125  Ibs.  abs.  and  2.5  cu.  ft.   Assume 
r  =  0.51. 


/I  \  IA        /    1  \  !-4 

t)      =(0)       =2-64  = 


P 

125.0 

62.5 

31.7 

15.7 

7.8 

3.9 

V 

2.5 

6.6 

17.4 

46.0 

122.0 

323.0 

MEAN  EFFECTIVE  PRESSURE 


To  find  the  probable  mean  effective  pressure  for  an  engine 
on  which  the  cut-off  and  limiting  pressures  are  known  but  for 
which  the  clearance  and  compression  are  not  known  as  the  pre- 


220 


HEAT  ENGINEERING 


liminary  dimensions  have  not  been  found,  the  best  procedure  is 
to  assume  zero  clearance  and  zero  compression  and  to  allow  for 
the  effect  of  these  by  empirical  constants.  The  card  abcde, 
Fig.  98,  shows  the  theoretical  card  in  this  case.  If  r  is  the 
apparent  ratio  of  expansion,  ab  will  equal  D/r  if  ed  is  assumed 
equal  to  D.  The  area  is  given  by 


D 


D 


area  =  -pi+~pi 


D_ 
D 


area       pi  , 
m.e.p.  =  -p-  =  -  [  1  +  loge  r]  -  pb 


(66) 


This  is  the  formula  by  which  the  mean  height  may  be  found  if 
the  line  is  assumed  to  be  the  rectangular  hyperbola.  For  the 
line  pvn  =  const. 


m.e.D.  =  - 


n  — 


for  all  values  of  n  except  1. 

If  clearance  is  added  to  the  diagram  as  shown  in  the  right-hand 
Fig.  98,  the  expansion  line  will  be  raised  and  for  that  reason  the 


g 

FIG.  98. — Theoretical   indicator   cards,    showing   effect   of   clearance   and 

compression. 

area  will  be  increased.  On  the  other  hand,  the  presence  of  com- 
pression in  the  cylinder  reduces  this  area.  The  card  is  then 
abfcdgh.  From  a  number  of  cards  drawn  with  different  clear- 
ances, pressures  and  cut-offs  the  net  result  of  these  two  effects 
was  to  decrease  the  area  or  m.e.p.  by  2  per  cent.  The  variation 
found  was  from  3  per  cent,  to  1  per  cent. 

REAL  AND  APPARENT  RATIO  OF  EXPANSION 

In  Fig.  98,  D/on  is  the  apparent  ratio  of  expansion  while  D/om 
is  the  real  ratio  of  expansion.     The  first  is  called  r  and  it  is  the 


THE  STEAM  ENGINE 


221 


reciprocal  of  the  cut-off.     If  the  clearance  is  ID,  the  real  ratio 
of  expansion  is 


Tr    = 


1  +  Ir 


The  actual  card  ab'c'de'f,  Fig.  99,  differs  from  the  theoretical 
card  abcdef  at  several  points.  In  most  cases  the  steam  is 
throttled  or  wire-drawn  during  admission.  This  causes  the  line 
ab  in  high-speed  engines  to  drop  about  10  per  cent,  in  pressure 
although  in  Corliss  engines  the  line  is  nearly  horizontal.  More- 
over, with  slide  valves,  the  slow  closing  of  the  valve  causes  further 
throttling  and  causes  the  corner  to  be  rounded.  The  expansion 
line  be  is  of  the  form  pvn  =  const.,  the  value  of  n  depending  on 


FIG.  99. — Actual  and  theoretical  cards  compared. 

the  quality  at  6.  The  release  takes  place  at  cr  before  the  end  of 
the  stroke  giving  the  line  c'd.  The  back  pressure  line  should  be 
that  assumed,  as  the  actual  value  is  taken  in  theory.  This  is  from 
J£  to  2  Ibs.  above  the  pressure  in  the  region  into  which  exhaust 
takes  place.  At  the  end  of  exhaust  the  back  pressure  line  will 
rise  due  to  the  slow  closing  of  the  exhaust  valve  so  that  the  point 
of  compression  ef  is  raised  above  e.  This  gives  the  line  e'f 
distinct  from  ef.  In  actual  cards  such  a  point  as  g  is  considered 
as  the  point  of  compression  but,  as  was  pointed  out  in  the  loga- 
rithmic diagrams,  this  point  is  not  on  the  compression  line. 
The  ratio  of  the  actual  area  to  the  theoretical  area  is  known  as 
the  diagram  factor. 


Area  (db'c'de'f) 
Area  (abcdef) 


=  diagram  factor. 


222 


HEAT  ENGINEERING 


The  value  of  this  for  single  cylinder  engines  is  about  0.90.  The 
actual  m.e.p.  is  then  given  by 

m.e.p.  =  0.90  X  0.9s[y  (1  +  log,  r)  -  pb]  (68) 

If  the  two  factors  are  combined  in  a  single  factor  known  as  the 
combined  factor,  this  may  be  found  from  the  results  of  an  engine 
test  by  the  following  formula: 

,  actual  m.e.p. 

combined  diagram  factor  =  -r —. — ; — - — 

theoretical  m.e.p. 


i.h.p.  X  33000 
2FLN 


X 


*  10000 

I    8000 

I 

£ 

g    6000 

o? 

|   4000 


Tests  give  the  value  of  this  to  be  about  0.85  for  high-speed  engines. 
To  find  the  value  of  the  m.e.p.  it  is  necessary  to  assume  or 

know  the  values  of  p\t  pb  and  r.  For  ordinary  single  cylinders  the 

value  of  pi  is  usually  90  Ibs. 
gauge  to  125  Ibs.  gauge  un- 
less it  is  desired  to  get  great 
power  or  force  from  a  rela- 
tively small  cylinder  when  a 
much  higher  pressure  is  used. 
If  high  pressure  is  used  cut- 
off must  be  made  very  early 
so  that  the  expansion  of  the 
steam  may  be  utilized  com- 
pletely. This,  however, 

FIG.  lOO.-Steam  consumption  curves.     CaUS6S  the  Percentage  effect 

of  initial  condensation  to  be 

felt  although  the  higher  temperature  range  gives  a  higher  theo- 
retical efficiency.  To  reduce  this  loss  a  later  cut-off  is  used  giving 
an  excessive  loss  due  to  free  expansion.  This  same  reasoning 
holds  for  the  determination  of  the  best  value  of  r.  If  r  is  large 
the  steam  is  used  to  advantage  expansively  but  the  percentage 
effect  of  initial  condensation  is  great  while  for  a  smaller  value 
of  r  this  condensation  effect  is  smaller  but  there  is  a  loss  from 
free  expansion. 

If  the  results  of  a  test  are  plotted  with  indicated  horse-power 
as  abscissae  and  total  weight  of  steam  per  hour  as  ordinates,  the 
most  efficient  point  is  found  at  the  point  of  tangency  of  a  straight 
line  from  the  origin  as  this  gives  the  smallest  ratio  of  ordinate 


150 
I.H.P. 


THE  STEAM  ENGINE 


223 


to  abscissa  which  is  the  steam  per  horse-power  hour.  In  Fig. 
100  this  relation  is  seen  for  the  curve  AB.  If  the  ratios  of. 
ordinates  to  abscissae  are  found  for  different  points,  the  steam 
consumption  curve  CD  is  found  on  which  F  is  the  best  point. 

Heck  proposes  that  the  weight  of  steam  per  hour  be  divided 
by  the  displacement  per  hour  giving  the  weight  of  steam  used  per 
cubic  foot  of  displacement.  In  this  case  the  line  AB  would  take 
the  same  form  because  the  displacement  per  hour  which  is 
120FLN  . 


144 


is  a  constant  for  Fig.  100. 

F  =  area  of  piston  in  square  inches. 

L  =  stroke  in  feet. 

N  =  revolutions  per  minute. 


Mean  Effective  Pres.suj'e 


FIG.  101.  —  Curves  of  steam  per  cu.  ft.  of  displacement  for  different  mean 

effective  pressures. 

This  quantity  is  called  m/,  weight  per  cubic  foot  of  displacement. 
wt.  per  hr. 


Now 


mf  = 


=  const< 


144 


from  which       pm  = 


.  _  2PmFLN 

i.h.p.  - 


33000  i.h.p. 
2FLN 


const.  X  i.h.p. 


(70) 


In  Fig.  101  the  line  AB  drawn  with  pm  and  ra/  as  coordinates 
is  the  same  curve  as  used  before  with  a  new  scale.     The  weight 


224 


HEAT  ENGINEERING 


of  steam  per  cubic  foot  of  piston  displacement  from  the  indi- 
cator card,  Fig.  102,  when  the  missing  quantity  is  m.q.,  is 


9 


1 


V   b 


1  -  m.q. 


(rv"b)(l-m.q.) 


-— c  approximately.    (71) 


FIG.  102. — Card  for  the  computation  of  steam  weight. 
Now  the  steam  per  indicated  horse-power  hour  is 

13750m/ 


or 


N  X  60 


2pmFLN 
33000 

Ml 
mf  =  13750^ 


(72) 


(73) 


That  is,  straight  lines  from  the  origin  of  Fig.  101  are  lines  of 
constant  steam  consumption;  the  consumption  being  equal  to 

13,750  times  the  ratio  —  or  the  inclination  of  the  lines. 
pm 

The  theoretical  amount  of  steam  per  cubic  foot  of  cylinder 
volume  is  l 

m<  =  ™»l  =  T  =  kr  (74> 

since  the  volume  at  cut-off  is  1/r  and  the  specific  weight  of  steam 

m0  =  a  +  bp  (75) 


is 


if  a  is  neglected  because  of  small  value.     This  results  from 


THE  STEAM  ENGINE 


225 


plotting  the  specific  weights  and  pressures  of  the  steam  tables 
as  shown  in  Fig.  103.  This  shows  clearly  that  the  value  of  m0 
is  practically  bp. 

Now  pm  =  ^  [1  -f  loge  r]  -  pb 


P 


[1  +  log.  rj -&•£-* 


(76) 


In  other  words  Mi  decreases  as  the  expansion  increases  since 
loge  r  increases  faster  than  r  — 


175 


150 


& 

3 

a  100 

£ 
P 

• 
• 

£     75 

0-1 

a 

!     50 


0.05 


0.10          0.15          0.20          0.25          0.30         0.35 
Weight  of  1  Cu.  Ft.  of  Saturated  Steam  in  Lbs. 


0.40 


FIG.  103. — Curve  of  relation  between  pressure  and  weight  of  1  cu.  ft.  of 

steam. 

The  amount  of  initial  condensation  per  cubic  foot  of  cylinder 
displacement  varies  slightly  with  r;  that  is,  as  pm  increases 
the  condensation  increases  slightly.  CD  will  represent  the 
initial  condensation  in  Fig.  101.  The  distance  from  AB  to  DC 
will  be  the  amount  shown  by  equation  (74).  From  this  figure 
Mi  can  be  found  as  well  as  from  (76). 

15 


226  HEAT  ENGINEERING 

The  best  point  is  usually  found  at  about  J4  or  %  cut-off  with 
single  cylinders  of  the  ordinary  form. 

While  discussing  steam  consumption  it  will  be  well  to  call 
attention  to  the  Willans  straight  line  law  for  the  steam  con- 
sumption of  throttle-governed  engines.  For  a  given  ratio  of 
expansion  and  back  pressure 

pm  =  cpi  -  pb 

2(cPl  -  Pb)FLN 
i.h.p.  =         -33000" 

2LFNXW      fcpi      2LFN      60fcr33000  i.h.p.    , 
Mtotal=       -144--X—      -344- X—[     2FLN      + 

=  k'  i.h.p.  +  k"pb  =  a  i.h.p.  +  b  (77) 

or,  the  total  weight  of  steam  per  hour  with  a  throttle  valve 
engine  varies  with  the  horse-power  on  a  straight  line. 


STUMPF  ENGINE 

Although  in  the  ordinary  engine  the  best  point  of  cut-off  seems 
to  be  about  one-fourth  stroke,  in  a  single  cylinder  engine  developed 
by  Prof.  Stumpf  in  1910  the  cut-off  is  carried  very  early  since 
the  initial  condensation  is  reduced  by  the  arrangement  of  the 
engine.  This  is  the  uniflow  (unidirectional  flow)  or  straight 
flow  engine  shown  in  Fig.  104.  This  was  invented  independently 
by  Stumpf  although  a  similar  idea  had  been  developed  by 
Bowen  Eaton  in  1857,  by  E.  Roberts  in  1874  and  by  L.  J.  Todd 
in  1885.  In  this  engine  steam  is  allowed  to  enter  from  the 
valve  A  behind  a  long  piston  B  and  forces  it  to  the  right.  After 
the  piston  moves  a  short  distance  to  the  right,  steam  is  cut  off  by 
the  double  beat  valve  and  the  steam  expands  to  nine-tenth  of  the 
stroke  when  the  piston  overrides  ports  C,  cut  in  the  cylinder  barrel, 
allowing  steam  to  exhaust  through  D  to  the  condenser  or  atmos- 
phere. The  area  thus  available  is  so  great  that  the  pressure 
rapidly  falls  and  when  the  piston  returns  and  covers  this  the 
steam  has  dropped  to  the  pressure  of  the  condenser  or  atmosphere. 
The  compression  then  begins  and,  by  properly  selecting  the 
clearance,  the  final  pressure  is  made  equal  or  nearly  equal  to  the 
boiler  pressure.  This  clearance  depends  on  the  back  pressure 
and;  if  this  is  liable  to  change  very  much,  arrangements  are  made 
to  vary  the  compression  or  clearance.  Thus,  with  a  condensing 


THE  STEAM  ENGINE 


227 


cylinder,  when  it  is  necessary  to  start  at  atmospheric  pressure 
until  the  air  pump  connected  with  the  engine  has  produced  the 
proper  vacuum,  an  auxiliary  clearance  volume  is  connected  to 
the  cylinder  and  is  controlled  by  a  valve  so  that  it  may  be  con- 
nected when  necessary. 

As  the  steam  expands  in  the  cylinder,  it  becomes  wet  due  to 
the  work  done,  as  is  known  from  the  discussions  of  the  adiabatic 
for  steam.  The  supply  steam  in  the  hollow  head  D,  however, 


FIG.  104. — Cylinder  of  Stumpf  engine  with  indicator  card. 

warms  the  steam  in  the  cylinder  in  contact  with  the  head  and 
keeps  it  dry  or  superheats  it  as  the  pressure  falls.  As  the  steam 
is  discharged  outward  at  the  end  of  the  stroke  the  steam  next 
to  the  head  expands,  driving  the  wet  steam  before  it,  and  leaves 
the  cylinder  practically  full  of  dry  steam.  As  the  piston  returns 
the  steam  compressed  is  superheated  as  it  starts  from  a  dry  or 
even  superheated  condition.  This  steam  gives  up  heat  to  the 
piston  head  so  that  the  steam  entering  the  cylinder  meets  a 
piston  surface  that  is  warm  while  the  head  and  walls  are  heated 


228 


HEAT  ENGINEERING 


to  the  temperature  of  the  incoming  steam  on  one  side  and  to  a 
temperature  higher  than  this  by  the  superheated  steam  on  the 
other  side.  The  part  of  the  cylinder  near  the  center  which  is  in 

contact  with  the  low  pres- 
sure or  exhaust  steam  is 
cut  off  by  the  piston  from 
contact  with  the  fresh  steam. 
The  curves  of  Fig.  105  made 
by  Stumpf  show  how  the 
wall  temperature  (a)  varies 
along  the  cylinder  of  a  con- 
densing engine  compared 
with  the  temperature  (6)  of 
the  saturated  steam  corre- 
sponding to  the  pressure 
when  the  piston  is  at  various 


FIG.  105. — Curve  showing  temperature 
of  wall  and  saturated  steam  for  different 
positions  of  piston  as  found  by  Stumpf. 


points.     The  fact  that  the 
metal  line  ig  aboye  the  satu. 

ration    line    indicates    that 
superheated  steam  must  be 

present  and  this  is  only  possible  on  compression.  The  cool 
central  portion  of  the  cylinder  is  a  good  feature  structurally 
for  lubrication  as  at  this  location  the  piston  is  moving  at  its 


t  20000 
n.  20  Ibs. 
94 


15000 

E   15  Ibs 


10000 
10  Ibs, 


«:     5000 
3    5  Ibs. 


10* 


20*  30*  40 

Mean  Effective  Pressure 


60* 


FIG.  106. — Steam  consumption  and  heat  per  h.p.  hr.  curves  of  Stumpf 

engine. 

highest  speed.  Stumpf  claims  that  the  steam  in  this  engine 
does  not  enter  in  a  disturbed  condition  and  that  it  flows 
always  toward  the  exhaust.  This,  of  course,  cannot  be  true  for, 


THE  STEAM  ENGINE  229 

until  release,  the  action  is  the  same  as  in  any  engine,  but,  at 
release,  the  steam  which  has  been  dried  or  superheated  by  the 
jacketed  head  drives  out  the  wet  steam,  whereas,  in  the  ordi- 
nary return  flow  engine,  this  steam  is  the  first  to  leave.  The 
distribution  of  the  exhaust  around  the  circumference  gives  an 
undisturbed  discharge  although  it  tends  to  remove  any  moisture 
from  the  piston  face.  The  jacket  head  being  in  the  steam  supply 
is  a  bad  feature  as  this  would  introduce  wet  steam  into  the 
cylinder  unless  superheated  steam  is  used.  The  effect  then 
would  be  to  reduce  the  amount  of  superheat  only  and  in  one  test 
this  amounted  to  54°  F.  or  27  B.t.u.  per  pound  of  steam.  The 
chambers  E  and  F  are  jackets  for  the  barrel  and  the  steam 
gradually  passes  from  one  to  the  other  and  with  this  a  drop  in 
pressure  causes  a  drop  in  temperature.  To  show  the  efficiency 
of  this  engine,  Fig.  106  is  presented.  This  is  from  a  test  of  a 
300  h.p.  engine  tested  with  saturated  steam  and  superheated 
steam  of  580°  F.  at  135  Ibs.  absolute  pressure.  The  results  are 
plotted  for  pounds  per  horse-power  hour  and  actual  B.t.u.  per 
horse-power  hour  chargeable  to  the  engine  plotted  against 
m.e.p.  They  show  the  closeness  of  results  with  saturated  and 
superheated  steam,  the  slight  variation  with  great  change  in  load 
and  values,  which  are  ordinarily  obtained  with  multiple  expan- 
sion engines.  This  engine  illustrates  the  great  value  of  expansion 
when  the  effects  of  initial  condensation  can  be  eliminated. 

SIZE  OF  ENGINES 

To  find  the  probable  size  of  an  engine  to  deliver  a  given  horse- 
power, the  formula  for  horse-power  is  used 

b.h.p.          2pFLN 


mech.  eff."  33000 


p  =  m.e.p.  in  pounds  per  square  inch. 

F  =  area  of  piston  in  square  inches. 

L  =  stroke  in  feet. 

N  =  revolutions  per  minute. 

In  this  equation  p  has  been  determined  from  the  formula 
(68)  after  pi,  pb  and  r  have  been  assumed.  N  is  fixed  by  the 
purpose  for  which  the  engine  is  to  be  used  or  by  the  valve  gear. 
For  Corliss  engines,  N  varies  from  60  to  120.  For  connected 
valve  gears,  N  may  be  as  high  as  450.  The  allowable  piston 


230  HEAT  ENGINEERING 

speed,  2LN,  then  fixes  L.  This  varies  from  250  in  small  engines 
to  1000  in  larger  engines.  The  only  unknown  is  now  F  by  which 
the  diameter  is  fixed. 

To  apply  this,  suppose  it  is  desired  to  find  the  size  of  an  engine 
to  drive  a  125  kw.  generator. 


0.746  XOO  X 


(79) 


This  number,  1.6,  is  an  important  average  number  to  keep  in 
mind  for  rapid  computations.  The  following  assumptions  will 
be  made:  pi  =  125  Ibs.  abs.,  pb  —  17  Ibs.  abs.,  r  =  4,  2LN  = 
500,  N  =  275. 

m.e.p.  =  0.90  X   0.98  [^  (1  +  2.3  X  0.602)  -  17J 
=  50.9  Ibs.  per  square  inch 

0.91  ft.  =  10.9  in. 


i.h.p.  X  33000  _    200  X  33000 

~  = 


_ 
2pLN    ~  2  X  50.9  X  0.91  X  275 


The  size  of  cylinder  necessary  would  therefore  be  18  in.  di- 
ameter X  12-in.  stroke,  the  somewhat  unusual  ratio  of  diameter 
to  stroke  being  caused  by  a  desire  to  keep  the  piston  speed  low. 

The  effect  of  clearance  on  an  engine  is  largely  dependent  on 
the  amount  of  surface  exposed.  The  volumetric  clearance  does 
not  seem  to  have  much  effect  on  the  efficiency.  Tests  have 
been  made  on  an  engine  with  varying  amounts  of  clearance  and 
the  results  have  shown  little  change  in  steam  consumption. 
The  effect  of  clearance  surface  as  shown  by  Heinrich  has  been 
mentioned  on  p.  209.  General  practice,  however,  is  to  cut  the 
clearance  to  a  minimum  so  as  to  increase  the  area  of  the  indicator 
card  for  a  given  displacement  and  to  make  the  steam  loss  due  to 
improper  compression  with  different  cut-offs  as  small  as  possible. 

BEST  POINT  OF  COMPRESSION 

The  effect  of  compression  is  very  slight  as  is  shown  theoretically 
and  by  the  tests  of  Heinrich  but  to  determine  the  best  point 


THE  STEAM  ENGINE 


231 


theoretically  several  methods  may  be  used.  The  amount  of 
compression  to  be  used  with  any  given  cut-off  may  be  found  by 
the  following  construction  of  Stumpf.  In  Fig.  107  the  initial 
pressure  is  pi,  and  the  back  pressure  pb,  the  curves  are  rectangular 
hyperbolae.  With  the  dimensions  given  on  the  card, 


Area  =  Pl  -  +  PlD  [I  +  -]  loge 


-  Dpb[l  -  x]  - 


pbD[l  +  x]  loge 


It  is  desired  to  eliminate  x  in  terms  of  the  volume  of  steam  taken 
from  the  boiler,  that  is,  x"D. 


xD 


FIG.  107. — Stumpf  s  method  of  finding  compression  point. 
Hence       Dpb(l  -  x)  =  Dpb[l  +  I]  -  Dpb[l  +  x] 


Dpb[l  +  1]-  DPl(l  +  I  -  x"] 


and 
pbD(l  +  x)  log 


-  PlD(l+  ~- 


area      pi  r7        1"] 

m.e.p  =  -p-    •fc.-f.p,^  +-J 


1  +  I 


.   71 
+  I] 


4- 


+  ~—  **]  - 


+     - 


JPi 


Suppose  now  that  the  volume,  x"D,  of  working  steam  re- 
mains constant  and  there  is  a  change  in  the  point  of  cut-off. 


232 


HEAT  ENGINEERING 


The  relation  which  gives  maximum  work  under  this  change  with 
constant  volume  of  working  steam  is  given  by  equating  the 
derivative  to  zero. 


d(m.e.p.) 


d    - 


l 


l+l 


l+l^     i  1 

+Pl~Pl\Oge-      —j—  -p^l+'-x") 


I  +--  -  x" 


Pl 


I    H X" 

r  Pi 

I  Pb 


l  +  X 


l+l      l+x 
1  ,  / 


(80) 


or  the  point  of  cut-off  divides  a  line  from  the  end  of  the  card  to 
the  axis  of  volume  in  the  same  proportion  as  the  clearance  di- 
vides a  line  to  the  axis  of  volume  from  the  point  of  compression. 


FIG.  108. — Webb's  method  of  finding  compression  point. 

Another  method  of  finding  the  point  of  compression  as  given 
by  Webb  in  the  American  Machinist  for  1890  is  to  carry  out 
the  expansion  line  to  the  back  pressure  line  as  in  Fig.  108  and 
then  lay  off  CB  so  that  the  area  ABCD  is  equal  to  area  EFG. 
The  point  C  is  then  taken  as  the  end  of  compression.  The  basis 
of  this  construction  is  the  fact  that,  by  changing  the  area 
ABCD  to  EFG,  the  card  CBEFGH  will  have  the  same  area  as 
ABEGHC  and  will  use  the  same  weight  of  steam  since  this  is 
proportional  to  the  line  CB.  The  card  CBEFGH  has  complete 


THE  STEAM  ENGINE  233 

expansion  and  must  have  complete  compression  to  overcome  the 
effect  of  clearance.  For  instance,  if  the  line  H'C'  were  used  for 
the  compression  line,  the  area  C'CHH'  would  have  to  be  saved 
to  make  up  for  the  additional  steam  CCr  since 

C'CHH'      CBEFGH 
C'C  CB 

But  the  area  JC'C  is  not  obtained  and  hence  the  gain  of  area 
JCHH'  is  not  proportional  to  the  increase  in  the  steam  quan- 
tity C'C. 

If  the  line  were  H"C"K,  then  the  saving  in  steam  would  be 
CC"  but  the  lost  work  would  be  H"HCKC".  This  is  greater 
than  the  proportional  reduction  H"HCC"  by  the  area  CKC" 
and  hence  this  does  not  pay. 

Figs.  107  and  108  are  for  the  same  conditions  and  the  results 
are  practically  the  same. 

Of  course  the  constructions  above  are  true  if  the  steam  is  as- 
sumed dry  or  proportional  in  all  cases  to  the  length  of  the  line 
between  the  compression  and  expansion  lines. 

The  methods  of  Clark,  see  his  Steam  Engine,  p.  399,  and  of 
Ball,  A.S.M.E.,  xiv.,  p.  1067,  may  be  referred  to  by  the  student. 
Clark's  method  is  the  equivalent  of  the  two  methods  given 
above  although  the  results  are  given  in  the  form  of  a  table. 

Having  the  compression  desired,  the  pressures,  the  cut-off, 
release  and  clearance  from  the  design  of  the  cylinder,  the  probable 
card  for  an  engine  may  be  drawn  and  studied. 

SPEED 

The  speed  of  an  engine  affects  its  efficiency  in  changing  the 
amount  of  condensation.  As  this  varies  inversely  as  the  cube 
root  of  the  number  of  revolutions  per  minute  it  would  naturally 
be  expected  that  high-speed  engines  would  be  the  more  efficient. 
That  this  is  not  so  is  due  to  the  fact  that  the  term  s  varies 
inversely  as  the  linear  dimension  of  the  engine  and  hence  this 
term  is  so  small  for  large  engines  that  it  overbalances  the  effect 
of  the  slower  speed.  For  instance,  if  the  piston  speed,  2LN,  is 
fixed,  the  initial  condensation  will  vary  approximately  inversely 
as  the  cube  root  of  N  and  also  inversely  as  the  square  root  of 
L.  Hence  it  would  pay  to  make  L  large  and  N  small.  This  is 
the  actual  result  in  practice.  The  large,  slow-speed  pumping 


234 


HEAT  ENGINEERING 


engines   represent   the   best   type   of   engines   using   saturated 
steam. 

The  effect  of  superheat  has  been  explained  and  results  given 
at  the  end  of  Chapter  II  show  that  this  materially  affects  the 


condensation  and  increases  the  efficiency  a  greater  amount  than 
that  estimated  by  theory. 

The  effect  of  the  jacket  on  a  cylinder  is  a  debated  question. 
The  results  shown  by  the  committee  of  the  Institution  of  Me- 


THE  STEAM  ENGINE  235 

chanical  Engineers  of  Great  Britain  indicate  a  distinct  gain. 
Other  tests  made  on  large  engines  show  no  gain  and,  in  a  few 
cases,  a  loss.  Of  course  the  total  steam  used  by  the  engine  is 
always  considered  in  these  cases.  In  most  instances  the  important 
saving  has  been  found  on  small  engines  of  less  than  300  h.p. 
For  large  engines  the  results  will  vary.  Ten  to  15  per  cent, 
may  be  saved  on  engines  of  200  or  300  h.p.  One  form  of  jacket 
which  has  bee.n  used  on  the  type  of  semiportable  engine  known 
as  the  locomobile  is  of  value.  In  this  unit  the  engine  is  mounted 
on  top  of  the  boiler,  Fig.  109,  and  the  exhaust  gases  from  the 
boiler  pass  around  the  cylinder,  thus  heating  it  to  a  high  tempera- 
ture. These  units  have  the  condensation  reduced  to  such  a  de- 
gree by  this  jacket  combined  with  the  use  of  superheated  steam 
that  they  give  a  brake  horse-power  hour  on  10^  Ibs.  of  steam 
1.25  Ibs.  of  coal  or  210  B.t.u. 

Effect  of  regulating  by  throttling  steam  and  changing  cut-off 
may  be  seen  when  the  formula  (53)  is  noted.  In  this  the 
quantity  s  changes  slightly  with  the  cut-off  and  so  there  is  a  slight 
increase  in  the  steam  due  to  this.  With  throttling,  however, 
there  is  an  amount  of  available  energy  lost  apparently.  When  it 
is  remembered  that  this  energy  is  used  in  changing  the  quality 
of  the  steam  which  becomes  drier  or  superheated,  it  is  seen  that 
when  the  steam  enters  the  cylinder  it  is  in  such  a  condition 
that  the  initial  condensation  is  less  and  hence  the  loss  due  to 
this  effect  is  smaller.  So  great  is  this  reheating  effect  with 
throttling  that,  even  with  this  steam  of  reduced  availability  the 
steam  consumption  is  as  low  as  with  engines  in  which  the  cut-off 
is  varied.  A  number  of  years  ago  the  author  experimented  on  a 
small  engine,  controlling  it  by  automatic  cut-off  and  by  throttle 
governing,  and  both  curves,  similar  to  CZ>,  Fig.  100,  were  coin- 
cident throughout  their  range.  The  curves  will  indicate  the 
steam  for  either  of  these  methods  and  also  how  the  consumption 
varies  with  the  load. 

TOPICS 

Topic  1. — Sketch  the  Rankine  cycle  of  the  steam  engine  with  complete 
expansion  and  compression  and  explain  the  difference  between  this  and  the 
Carnot  cycle.  Explain  the  variation  in  quality  on  the  various  lines  in  these 
two  cycles.  Show  that  clearance  has  no  effect  in  this  case  when  initial 
condensation  is  not  present. 

Topic  2. — Sketch  the  Rankine  or  Clausius  cycles  with  complete  expansion 
and  incomplete  expansion  on  the  p-v  and  T-S  planes  and  derive  the  various 


236  HEAT  ENGINEERING 

expressions  for  efficiency.  Explain  why  the  cards  may  be  drawn  with  no 
compression  lines. 

Topic  3. — Sketch  the  T-s  diagram  of  a  Rankine  cycle  with  incomplete 
expansion  and  discuss  the  value  of  increasing  the  steam  pressure,  vacuum  or 
amount  of  superheat.  Does  the  increase  of  vacuum  always  pay?  Why? 

Topic  4. — Sketch  and  explain  action  of  the  Barrus  calorimeter.  Give 
the  method  of  computing  results.  Explain  what  is  meant  by  a  dry  test,  how 
it  is  made  and  for  what  it  is  used  ?  What  is  a  separating  calorimeter  ?  What 
is  the  formula  for  x  when  this  instrument  is  used.  What  is  an  electric  calo- 
rimeter? What  are  constant  immersion  thermometers?  Why  are  they 
used?  What  is  the  purpose  of  the  upper  thermometer  of  the  Barrus 
calorimeter? 

TopicS. — What  is  Hirn's  analysis?  For  what  is  it  used?  What  observa- 
tions are  made?  Explain  how  the  quantities  M0  and  M  are  found.  Ex- 
plain how  a?i,  xz,  and  x3  are  found.  Explain  how  AUi,  AUz,  AU3  and  AU* 
are  found.  Explain  how  Qi  and  Q2  are  found. 

Topic  6. — Given  the  average  indicator  card,  show  how  the  quantities 
AW a,  AWb,  AWC  and  AWd  are  found.  Give  method  of  changing  from 
square  inches  to  B.t.u.  Having  the  U's,  Q's  and  AW's  explain  how  Qa,  Qb, 
Qc  and  Qd  are  found.  What  check  equation  may  be  used  for  these  quanti- 
ties? How  is  the  new  quantity  for  this  check  equation  found?  What  differ- 
ence does  a  jacket  make  in  these  equations? 

Topic  7. — What  is  the  temperature-entropy  analysis?  What  readings  are 
necessary?  Explain  how  the  curves  of  the  quadrants  are  constructed  and 
sketch  the  diagram.  How  is  the  average  indicator  card  found  and  how  is 
this  transferred  to  the  p-v  quadrant  of  the  diagram. 

Topic  8. — Explain  how  to  transfer  the  diagram  from  the  p-v  quadrant  to 
the  T-s  quadrant.  Explain  the  method  of  finding  the  various  losses.  Which 
lines  on  the  diagram  are  true?  Why  can  the  diagram  be  used  for  other 
lines? 

Topic  9. — What  is  initial  condensation?  How  large  may  it  be?  To  what 
is  this  due?  What  is  the  missing  quantity?  In  what  units  is  it  found? 
Explain  how  it  may  be  found  from  a  test.  Explain  what  is  meant  by  the 
terms  of  the  two  formulae  below : 


mq 


30  V7 


-  mq 

0.27 


Topic  10.  —  Derive  the  formulae: 

M.  =  K'Fc~ 


V't 

Why  is  it  that  it  may  be  said  that  v"2  =  -  ?     What  is  Heck's  expression 

0.27      fsr 

fors?    Using  Heck's  formula  m.q.  =  ~j/=  \  -  discuss  the  effect  of  size, 

pe 


THE  STEAM  ENGINE  237 

speed,  pressure  and  cut-off  on  the  actual  condensation  and  on  the  percent- 
age condensation,  m.q. 

Topic  11. — Give  a  statement  of  the  work  of  Callendar  and  Nicolson, 
Hall,  Duchesne  and  Adams  on  the  effect  of  cylinder  walls.  How  do  Callendar 
and  Nicolson  plot  their  results  so  that  initial  condensation  may  be  found? 
What  do  their  results  show  in  regard  to  variation  of  steam  temperature  in  the 
cylinder,  and  in  a  small  hole  in  the  head  ?  What  is  the  nature  of  the  tempera- 
ture cycle  at  various  points  in  the  cylinder  length  and  at  various  depths? 
Which  has  the  greater  effect,  clearance  surface  or  clearance  volume?  When 
does  the  major  part  of  the  initial  condensation  take  place?  To  what  is  it 
due? 

Topic  12. — Explain  Clayton's  method  of  finding  the  quality  at  cut-off, 
constructing  the  diagram  from  the  indicator  card.  Show  how  to  find  the 
probable  steam  consumption  from  the  indicator  card.  Explain  how  leaks 
are  detected.  Explain  how  to  locate  the  events  of  the  stroke  properly. 

Topic  13. — What  values  of  n  are  to  be  expected  on  various  machines  on 
compression  and  expansion?  Explain  how  the  following  curves  may  be 
constructed  on  the  pv-plane:  (a)  rectangular  hyperbola,  (6)  adiabatic,  (c) 
constant  steam  weight  curve  and  (d)  pv1-05  =  const.  What  conclusions 
may  be  drawn  from  the  figure  in  the  book  showing  the  various  expansion 
curves? 

Topic  14. — Explain  how  to  construct  the  curves  pV  =  const,  and  pVn  = 
const,  through  a  given  point  p\V\.  Find  the  area  of  the  indicator  card  for 
each  of  these  to  a  volume  F2  and  pressure  p2  if  the  back  pressure  is  pi.  From 
this  find  the  formulae  for  the  m.e.p. 

Topic  16. — Explain  what  is  meant  by  the  real  and  apparent  ratio  of  expan- 
sion. Derive  a  formula  for  the  real  ratio  of  expansion  in  terms  of  the 
apparent  ratio,  r,  and  the  clearance,  I.  What  is  meant  by  diagram  factor? 
How  is  it  found?  What  value  does  it  have? 

Topic  16. — Sketch  the  curve  of  total  steam  consumption  and  show  how  to 
find  from  this  the  curve  of  steam  consumption  per  horse-power  hour.  Ex- 
plain why  this  curve  is  the  same  as  the  curve  between  ra/  and  pm.  What  is 
the  Willans  straight-line  law?  Prove  that  it  is  true. 

Topic  17. — What  is  the  Stumpf  engine?  Why  is  it  of  value?  Sketch  it. 
Sketch  curves  showing  the  temperatures  of  the  wall  and  saturated  steam  for 
various  positions  along  the  length  of  the  cylinder.  What  conclusion  may 
be  drawn  from  this?  Sketch  the  cards  from  this  engine.  How  is  compres- 
sion cared  for  when  high  back  pressure  exists  in  starting  a  condensing  Stumpf 
engine  with  a  direct-connected  air  pump? 

Topic  18. — Explain  how  to  find  the  indicated  horse-power  required  to 
produce  a  given  kilowatt  output  from  a  direct-connected  generator.  Show 
how  to  find  the  size  of  an  engine  to  produce  this  power. 

Topic  19. — Derive  the  formula  for  the  best  point  of  compression. 

Topic  20. — Sketch  the  constructions  for  the  best  point  of  compression  due 
to  Stumpf  and  to  Webb.  What  is  the  effect  of  speed?  Of  superheat?  Of 
jackets?  What  difference  is  found  between  governing  by  throttling  and  by 
automatic  cut-off? 

Topic  21. — Sketch  and  explain  particular  features  of  the  locomobile. 


238  HEAT  ENGINEERING 

PROBLEMS 

Problem  1. — Find  the  various  efficiencies  of  an  engine  with  pi  =  120  Ibs. 
gauge  pressure,  p2  =  30  Ibs.  gauge  pressure,  pl  =  2  Ibs.  gauge  pressure  if  x: 
=  0.99  and  35  Ibs.  of  steam  are  needed  per  i.h.p.-hr. 

Problem  2. — One  engine  uses  25  Ibs.  of  steam  per  i.h.p.-hr.  with  steam  at 
125  Ibs.  gauge  pressure  and  of  quality  0.995.  The  back  pressure  is  2  Ibs. 
by  gauge.  By  using  steam  of  200°  F.  superheat  the  steam  consumption  is 
reduced  to  22  Ibs.  per  i.h.p.-hr.  What  was  the  saving? 

Problem  3. — An  engine  uses  3500  Ibs.  of  steam  per  hour,  of  which  300  Ibs. 
is  used  in  the  jackets  and  200  Ibs.  in  the  receiver.  The  steam  supply  is  at 
150  Ibs.  gauge  pressure  with  175°  F.  superheat.  The  temperature  of  the 
return  from  the  jackets  is  320°  F.,  while  the  return  from  the  receiver  is  338° 
F.  and  the  hot  well  temperature  from  the  condensate  is  95°  F.  The  engine 
develops  250  h.p.  Find  the  B.t.u.  per  h.p.-min.  Find  the  actual  efficiency. 
If  the  mechanical  efficiency  of  the  pump  and  engine  combined  is  92  per  cent., 
what  is  the  duty  of  this  engine? 

Problem  4. — In  Fig.  68  assume  that  the  pressure  on  the  top  line  is  120 
Ibs.  absolute  and  on  the  lower  line  is  15  Ibs.  absolute.  Suppose  that  the  qual- 
ity on  the  top  line  is  0.98,  assuming  the  cycle  to  be  the  Rankine  cycle,  and 
that  it  varies  from  0.03  to  0.98  if  assumed  to  be  the  Carnot  cycle.  Find  the 
qualities  at  the  lower  corners.  Find  the  heats  on  the  four  lines. 

Problem  5. — In  a  Rankine  cycle  with  complete  expansion  the  pressure 
varies  from  125  Ibs.  gauge  to  0  Ibs.  gauge  with  xi  =  1.0.  Find  the  efficiencies, 
•ni,  1)2, 173.  Increase  the  upper  pressure  to  150  Ibs.  gauge  and,  leaving  the  other 
quantities  unchanged,  find  771,  772  and  173.  With  125  Ibs.  initial  gauge  pres- 
sure assume  the  quality  changes  to  160°  F.  superheat;  find  171,  772  and  773. 
Assume  the  back  pressure  is  changed  to  a  vacuum  of  27  in.  but  with  no 
change  in  other  conditions;  find  771, 772  and  773. 

Problem  6. — In  a  Rankine  cycle  with  incomplete  expansion  let  the  initial 
gauge  pressure  be  125  Ibs.,  the  pressure  at  the  end  of  expansion  20  Ibs.  gauge 
and  the  back  pressure  is  that  of  the  atmosphere.  If  x\  =  1.0  find  the  three 
efficiencies,  771, 772  and  773.  Change  the  initial  pressure  only  to  150  Ibs.  gauge 
and  find  the  efficiencies.  Change  the  initial  quality  to  160°  F.  superheat 
and  find  the  efficiencies.  Change  the  back  pressure  only  to  27  in.  and 
find  the  efficiencies. 

Problem  7. — The  following  results  were  obtained  from  a  test  of  an  engine : 

Size  of  engine  10  in.  X  14  in.  (neglect  rod) 

Time  of  test 60  min. 

Clearance 7  per  cent. 

Number  of  revolutions 15,000 

Steam  used 3003  Ibs. 

Average  gauge  pressure  at  throttle 112  Ibs.  per  sq.  in. 

Barometric  pressure 14.7  Ibs.  per  sq.  in. 

Average  quality  of  steam 0.99 

Average  temperature  of  condensate 135°  F. 

Average  temperature  of  water  leaving 120°  F. 

Average  temperature  of  water  entering 75°  F. 

Weight  of  condensing  water 58,100  Ibs. 


THE  STEAM  ENGINE  239 

Average  results  from  indicator  cards 

Point  of  admission 0.0%  of  stroke,. ...   55      Ibs.  abs.  per  sq.  in. 

Point  of  cut-off 33 . 0%  of  stroke, ....  114 . 7  Ibs.  abs.  per  sq.  in. 

Point  of  release. 93.0%  of  stroke,. ...   44.7  Ibs.  abs.  per  sq.  in. 

Point  of  compression 20.0%  of  stroke,. ...    14.7  Ibs.  abs.  per  sq.  in. 

Work  of  admission 3.22  sq.  in. 

Work  of  expansion 3.32  sq.  in. 

Work  of  exhaust —  0.70  sq.  in. 

Work  of  compression —  0.46  sq.  in. 

Make  Hirn's  analysis  from  the  above  data. 

Problem  8. — -In  the  analysis  above  the  following  coordinates  give  the 
positions  of  the  points  in  inches  on  a  4-in.  card  from  the  true  zero  of  pressure 
and  volume  with  a  spring  scale  of  50  Ibs.  per  inch. 

Points  Volume  Pressure  Points  Volume  Pressure 

1  0.28  2.48  10  4.28  0.29 

2  0.78  2.44  11  3.28  0.29 

3  1.28  2.36  12  2.28  0.29 

4  1.60  2.28  13  1.08  0.29 

5  2.28  1.60  14  0.78  0.41 

6  3.28  1.19  15  0.48  0.63 

7  3.78  0.94  16  0.28  1.08 

8  4.02  0.88  17  0.28  1.78 

9  4.18  0.60  18  0.28  2.30 

Plot  the  card  and  make  the  T-S  analysis.     Find  the  various  losses. 

Problem  9. — Find  the  probable  value  of  the  missing  quantity  for  the  engine 
given  in  Problem  8  by  the  various  formulae  of  this  chapter. 

Problem  10. — If  the  ratio  of  connecting  rod  to  crank  is  6  to  1  mark  the 
piston  positions  for  every  five-sixtieths  of  a  revolution  on  the  indicator  card 
constructed  from  the  data  of  Problems  7  and  8.  Also  find  crank  positions  of 
events  of  stroke.  Construct  the  Callendar-Nicolson  diagram  for  this  card 
and  find  the  probable  initial  condensation.  Express  this  as  a  percentage  of 
the  steam  given  in  Problem  7  and  compare  results  with  those  of  Problems 
7,  8  and  9. 

Problem  11. — Using  data  of  Problem  8  construct  table  and  diagram  used  in 
Clayton's  method.  Find  probable  initial  condensation  and  steam  consump- 
tion. Compare  this  with  other  results. 

Problem  12. — Find  the  real  and  apparent  ratios  of  expansion  of  the  card  of 
problems  7  and  8.  Find  the  diagram  factor.  Find  the  horse-power  de- 
veloped in  Problem  7.  Find  the  actual  steam  consumption. 

Problem  13. — Construct  a  curve  of  total  steam  consumption  and  from  it 
find  the  curve  of  steam  per  i.h.p.-hr. 

Problem  14. — Find  the  size  of  a  single-cylinder  non-condensing  engine  to 
drive  a  125-kw.  generator  with  initial  gauge  pressure  of  130  Ibs.,  cut-off  at 
0.3  stroke  and  a  back  pressure  of  2  Ibs.  gauge. 

Problem  15. — Construct  the  card  for  Problem  14  and  find  the  best  point 
of  compression  for  10  per  cent,  clearance  by  Stumpf's  method  and  Webb's 
method.  Release  is  at  95  per  cent,  of  stroke. 


CHAPTER  VI 


\ 


p.  c 


MULTIPLE  EXPANSION  ENGINES 

Multiple  expansion,  which  is  the  use  of  steam  in  one  cylinder 
after  another,  was  introduced  to  cut  down  the  wastes  in  steam 
engines.  After  its  introduction  it  was  seen  that  for  structural 
reasons  such  an  arrangement  is  of  value,  as  it  reduces  the  sizes 
of  parts  of  the  machine  and  gives  a  more  uniform  turning 
moment.  The  reason  for  the  reduction  in  waste  is  the  fact 
that  there  is  a  series  of  small  ranges  in  temperature;  that 

some  of  these  ranges  act  on 
small  surfaces  (i.e.,  in  the 
smaller  high-pressure  cylin- 
ders), and  lastly  that  heat  ab- 
stracted by  the  exhaust  steam 
from  the  upper  stages  may 
be  of  value  for  use  in  the  lower 
stages.  If  the  indicator  cards 
from  a  two  stage  or  compound 
engine  are  taken  as  shown  in 
Fig.  110,  these  cards  may  be 
used  as  shown  in  the  preced- 
ing chapter  to  study  the  ac- 
tion of  the  cylinder  walls,  to 
find  the  horse-power  and  to 
use  for  any  purpose  that  cards 
for  simple  engines  are  used.  However,  to  get  a  fuller  picture 
it  is  often  desired  to  construct  what  is  known  as  a  combined  dia- 
gram. To  form  this  the  cards  of  Fig.  110  from  a  10  and  20  X  24 
engine  are  divided  by  ordinates  at  regular  intervals,  say  10,  after 
laying  off  the  clearance  at  the  end  of  each  diagram.  The  extreme 
ends  of  the  diagrams  are  placed  on  a  new  axis  of  volume  and  the 
base  lines  with  their  ordinates  are  drawn  in  after  making  the  card 
lengths  proportional  to  the  volumes  swept  out  by  the  respective 
pistons.  In  most  cases  since  the  strokes  are  equal  these  are  pro- 
portional to  the  squares  of  the  diameters.  This  increases  or  de- 

240 


L. 

.. 

,/ 

\ 

>s 

p^ 

r  — 

\ 

\ 

•*^^. 

16* 

Spri 

ng 

7S 

FIG.  110. — Indicator  cards  from  high 
and  low  pressure  cylinders  of  com- 
pound engine. 


MULTIPLE  EXPANSION  ENGINES 


241 


FIG. 


111. — Combined  cards  for  com- 
pound engine. 


creases  all  horizontal  dimensions.  If  now  a  scale  be  assumed  for 
pressure  and  the  ordinates  from  the  atmospheric  pressure  line  are 
measured  on  the  cards  and  reduced  to  this  new  scale,  the  points 

may  be  plotted  on  the  new 
figure  and  the  cards  of  Fig. 
Ill  drawn.  These  are  both 
drawn  to  the  same  scales  of 
volume  and  of  pressure.  The 
diagram  is  known  as  the  com- 
bined diagram.  Since,  how- 
ever, the  weights  of  clearance 
steam  in  each  cylinder  are  not 
the  same,  although  the  work- 
ing steam  is  equal  on  each, 
these  figures  do  not  represent 
diagrams  for  the  same  total 
weight  of  steam  on  the  ex- 
pansion line.  For  this  reason 
the  expansion  line  of  one  cyl- 
inder does  not  pass  through 
that  of  the  other.  The  same  is  true  of  the  compression  lines. 
These  cards  show  in  this  figure  the  relative  amounts  of  work 
done  by  each  cylinder  and  the  amount  lost  due  to  the  drop  in 
pressure  between  the  two  cylinders.  That  these  diagrams  cut 
each  other  means  noth- 
ing since  the  diagrams 
are  not  simultaneous  for 
the  same  piston  positions. 
The  first  cylinder  is 
known  as  the  high-pres- 
sure cylinder  and  the 
second  as  the  low-pres- 
sure cylinder.  If  there 
are  three  stages,  known 
as  triple  expansion,  the 
cylinders  are  known  as 
high  pressure,  interme-  pIG  H2.— Combined  cards  from  Woolf  com- 

diate  pressure  and  low  pound  engine. 

pressure.      In    addition 

to  high-  and  low-  pressure  cylinders  there  are  first  and  second  in- 
termediate pressure  cylinders  in  the  quadruple  expansion  engine. 

16 


242  HEAT  ENGINEERING 

When  the  cylinders  and  cranks  are  so  arranged  that  when  the 
high-pressure  cylinder  is  ready  to  discharge  steam  the  low-pres- 
sure cylinder  is  ready  to  receive  it,  the  engine  is  known  as  a 
Woolf  compound  engine,  while  if  the  discharge  must  be  stored 
in  a  receiver  before  it  can  be  used,  the  arrangement  is  known  as 
a  receiver  engine.  At  times  several  cylinders  may  be  used  for 
any  stage,  say  low,  for  the  purpose  of  reducing  the  size.  For  the 
diagram  of  the  Woolf  engine  see  Fig.  1 12.  The  cut-off  in  the  high- 
pressure  cylinder  occurs  at  1  and  the  steam  expands  to  2.  At 
this  instant  the  steam  is  connected  to  the  low-pressure  cylinder 
in  which  the  pressure  is  that  shown  by  12  together  with  the  con- 
necting pipe  in  which  the  pressure  was  left  the  same  as  at  8. 
Hence  the  pressure  falls  from  2  to  3  in  the  high  and  rises  from  12 
to  13  in  the  low  and  from  that  of  8  to  that  of  3  or  13  in  the 


8      1 


FIG.  113. — Combined  cards  from  receiver  engine. 

receiver  or  connecting  pipe.  The  pressure  at  13  is  the  same  as 
that  at  3  except  for  a  drop  due  to  throttling  in  the  passage  through 
the  connecting  pipe.  As  the  high  pressure  forces  the  steam  out 
it  enters  the  low-pressure  cylinder  in  which  there  is  a  larger 
piston  moving  at  the  same  speed  as  that  of  the  high-pressure 
cylinder,  and  the  pressure  drops  gradually  due  to  this  increase  of 
volume.  This  finally  reaches  the  point  of  compression  on  the 
high  pressure  cylinder  which  corresponds  to  the  point  of  cut-off 
on  the  low-pressure  cylinder  in  many  cases.  The  pressure 
at  7  and  that  at  4  are  about  the  same.  The  line  from  4  to  5  is 
the  compression  and  from  5  to  1  there  is  admission.  The  steam 
in  the  low-pressure  cylinder  and  connecting  pipe  or  receiver  now 


MULTIPLE  EXPANSION  ENGINES  243 

expands  from  7  to  8  at  which  point  the  cut-off  occurs  in  the  low- 
pressure  cylinder.  From  8  to  12  the  events  on  the  low  are  the 
same  as  those  on  a  single  cylinder. 

The  combined  cards  from  a  receiver  engine  are  shown  in  Fig. 
113.  In  this  case  at  the  end  of  expansion  the  high-pressure 
cylinder  discharges  into  the  receiver  in  which  the  pressure  is  that 
at  11  if  cut-off  in  the  low  pressure  occurs  after  compression  on 
the  high-pressure  cylinder,  while  the  pressure  is  that  of  6  if  this 
compression  in  the  high  occurs  later  than  the  cut-off  in  the  low. 
In  either  case,  the  pressure  in  the  receiver  is  usually  lower  than 
that  at  2  and  there  is  a  drop  in  pressure.  The  low-pressure 
cylinder  is  not  in  a  position  to  take  steam  at  this  point  as  in  most 
cases  the  cranks  of  the  engines  are  at  right  angles  and  for  that 
reason  the  receiver  is  used.  The  amount  of  drop  2-3  not  only 
depends  on  the  pressure  in  the  receiver  but  also  on  the  relative 
volume  of  the  receiver  and  the  cylinder.  This  may  be  1  J£  times 
the. volume  of  the  high-pressure  cylinder.  As  the  piston  now 
returns  the  steam  is  compressed  into  the  clearance  space  and  the 
receiver  to  the  point  4.  Here  steam  enters  the  low-pressure 
cylinder  in  which  the  pressure  is  15.  There  is  a  drop  from  4  to 
5  in  the  receiver  and  high-pressure  cylinder  and  a  rise  from  15  to 
9  in  the  low.  The  line  3-4  is  a  rectangular  hyperbola  with  the 
origin  at  a  distance  of  the  receiver  volume  from  the  clearance 
line.  From  5  to  6  the  small  piston  at  the  middle  of  its  stroke  is 
moving  so  much  faster  than  the  large  piston  at  the  beginning  of 
its  stroke  that  the  volume  decreases  and  the  pressure  rises  at 
first.  It  then  usually  falls  before  6  is  reached,  9  agrees  with  5, 
and  10  with  6.  Cut-off  on  the  low  occurs  after  compression  on 
the  high  but  before  the  high  pressure  reaches  the  end  of  its  stroke. 
If  it  did  not  occur  sooner  than  this  there  would  be  a  second 
admission  of  steam  from  the  other  end  of  the  high-pressure 
cylinder.  The  other  points  are  clearly  seen. 

COMPUTATION  OF  CARDS  FOR  CONSTRUCTION 

To   compute   these  various   points   the   volumes  of   the   two 
cylinders  must  be  known,  with  the  clearances;  the  volume  of  the 
receiver  must  be  known,  and  finally  the  points  of  all  the  events 
of  each  card  and  the  initial  pressures. 
Let   Vr  =  receiver  volume. 

V  =  volume  of  cylinder  at  event  represented  by  sub- 
script including  clearance. 
p  with  subscript  =  pressure  at  point. 


244  HEAT  ENGINEERING 

Now  from  the  curve  on  Fig.  103  it  is  seen  that  m  =  kp  ap- 
proximately, hence  the  total  weight  of  saturated  steam  at  any 
point  is 

M  =  mV  =  kpV  (1) 

or  the  weight  of  steam  is  approximately  equal  to  the  product 
pV.  Hence  if  quantities  of  steam  of  different  volumes  and  at 
different  pressures  are  connected  the  resultant  pressure  mul- 
tiplied by  the  sum  of  the  volumes  must  be  equal  to  the  sum  of 
the  individual  products  of  pressure  and  volume,  or 


With  this  understanding,  the  following  equations  hold: 

i     ,,  ,  ,„, 

(known) 

yf'      (Pn  unknown)  (4) 

+  (In  terms  of  pi,)  (5) 


p,(Vt  +  Vr)  +  pnVu     (In  terms  of  pu  ;  since          ,  , 
F4  +  Vr  +  Fi5~       pi6V15  is  known) 

(7) 

°f 


ee  /ON 

Pl  = 

yr)      (In  terms  of  PiiJ  •"•  P"  is  known)  (10) 

/-,-,\ 

To  apply  these  a  table  is  first  computed  for  all  volumes  in  pro- 
portional numbers  if  not  in  actual  numbers,  and  from  this  table 
the  substitution  can  be  made  in  the  equations  above.  To  find 
the  relative  position  of  events  a  diagram  is  made  with  the  cranks 
as  shown  in  Fig.  114.  The  actual  position  could  be  found  by 
using  the  connecting  rods  or  if  infinite  rods  are  assumed  the  pro- 
jections may  be  used.  Thus  if  the  compression  is  to  occur  at 
0.1  stroke  on  the  high-pressure  cylinder,  the  diagram  for  an  in- 
finite connecting  rod  would  be  shown  in  Fig.  114,  with  a  crank 


MULTIPLE  EXPANSION  ENGINES 


245 


of  0.5.  From  this  it  is  seen  that  the  low-pressure  piston  will 
have  moved  0.2  of  its  stroke  when  compression  occurs  in  the 
high-pressure  cylinder.  The  high-pressure  piston  is  at  the  middle 
of  the  stroke  when  the  low-pressure  stroke  begins. 

Suppose  that  the  cylinders  have  a  ratio  of  1  to  4  and  the 
clearance  is  6  per  cent,  on  the  high  and  5  per  cent,  on  the  low. 
Suppose  that  the  receiver  volume  is  0.8  times  the  volume  of  the 
high-pressure  cylinder;  that  the  cut-off  occurs  at  0.33  stroke  in 
high-pressure  cylinder  and  0.45  stroke  in  low;  that  compression 
is  at  0.1  stroke  from  end  on  high  and  0.2  on  low  and  that  pi 
is  150  Ibs.  gauge  and  pb  is  2  Ibs.  absolute. 


A  L.P. 


FIG.  114. — Diagram  by  which  to  find  piston  position.     (Finite  or  infinite 

connecting  rod.) 

Call  the  volume  swept  out  by  the  high-pressure  cylinder  1  and 
using  Fig.  113: 


Vh    =  I 
Vi    =4 

Vr     =    0.8 

Vch  =  0.06  X  1  =  0.06 
Vcl  =  0.05  X  4  =  0.20 


Fi    =  0.33  +  0.06  =  0.39 

F2   =  1.00  +  0.06  =  1.06  =  F3 

F4   =  0.50  +  0.06  =  0.56  =  F5 

F6   =  0.10  +  0.06  =  0.16 

V7   =  0.06  =  F8 

V  9  -  0.20  =  Fi5. 

Fio  =  0.20  X  4  +  0.20  =  1.00 

Fii  =  0.45  X  4  +  0.20  =  2.00 

y12  =  4  +  0.20  =  4.20  =  Fig 

Fi4  =  4  X  0.20  +  0.20  =  1.00 


=  164.7  X 
60.6  X  1 


=  60.6 
p,i  X0.8 


P4    = 

P5    = 


(34.5 


1.06  +  0.8 
0.43p»)  1.86 


34.5 


=  47'2 


0.56  +  0.8 
(47.2  +  0.59piQ  1.36  +  10  X  0.2 
1.36  +  0.2 


=  43.2  +  0.514  pn 


246  HEAT  ENGINEERING 

1  00 
PIB  =  2  X  -^  =  10 

(43.2  +  0.514Pll)  1.56 
Pe  =  pio  =       0.16  +  0.8+1.00       =  34'4  +  °'41 

(34.4  +  0.41P11)(1.00  +  0.8) 

Pn  2.00  +  0.8  =  22'1  +  °'264 

22.1 

3° 


0736 

Pio  =  p6  =  46.7 
Pb  =  PQ  =  58.7 
pi  =  64.9 
2>3  =  47.4 

P7  =  46.7  x  §^  =  125      |;||f    *       ;    J-;  . 

2  00 
Pu  =  30  X  j^o  =  14.3 

This  method  could  be  used  for  triple  expansion  engines  or  for 
any  arrangement  of  cylinders.  The  main  point  to  remember 
is  to  reduce  the  equations  for  any  unknown  to  terms  of  one  un- 
known until  at  last  this  term  occurs  on  both  sides  of  the  equa- 
tion and  the  equation  gives  its  value. 

EQUIVALENT  WORK  DONE  BY  ONE  CYLINDER 

On  looking  at  the  cards  of  Figs.  Ill,  112  and  113  it  will  be 
seen  that  the  total  length  of  the  combined  diagram  represents 
the  volume  of  the  low-pressure  cylinder  and  if  the  intermediate 
lines  were  removed  the  card  would  appear  as  a  single  cylinder 
card  with  very  early  cut-off.  The  cylinder  would  be  of  the  same 
size  as  the  low-pressure  cylinder.  Hence  it  may  be  said  that  if 
the  same  initial  pressure  is  used  in  a  cylinder  of  the  size  of  the 
low-pressure  cylinder  of  a  multiple  expansion  engine  as  is  used 
in  the  high-pressure  cylinder,  and  if  this  has  the  same  total  ratio 
of  expansion,  then  it  will  develop  the  same  horse-power  as  the 
multiple  expansion  engine. 

It  is  well  to  note  what  differences  occur.  With  the  single 
cylinder  the  whole  clearance  surface  of  the  large  cylinder  is 
subject  to  the  full  temperature  range  which  would  cause  excessive 
condensation  unless  it  is  of  the  Stumpf  form  of  cylinder.  The 
large  cylinder  would  be  subject  to  full  high  pressure  not  only 
requiring  heavy  walls  for  the  cylinder  but  causing  the  rods,  pins 


MULTIPLE  EXPANSION  ENGINES 


247 


and  bearings  to  be  excessive.  The  cut-off  on  the  single  card  oc- 
curs at  about  one-tenth  stroke,  while  on  each  of  the  separate 
cards  the  division  of  the  expansion  caused  the  pressure  to  be 
continued  to  nearly  the  middle  of  the  stroke  on  a  smaller  area 
giving  a  more  gradual  curve  of  turning  moment. 

On  looking  at  the  computations  made  above  it  will  be  seen 
that  if  the  receiver  volume  had  been  made  very  large  there  would 
have  been  little  change  in  the  intermediate  lines  and  were  the 
receiver  infinite  in  volume  these  lines  would  be  horizontal.  Such 
an  assumption  is  usually  made  in  determining  the  preliminary 
relative  sizes  of  the  cylinders  of  a  multiple  expansion  engine  and 
in  addition  the  effect  of  clearance  and  compression  are  omitted 
in  this  work.  Although  in  practice  the  receiver  is  not  more 
than  1^2  times  the  volume  of  the  cylinder  which  discharges 
into  it,  it  is  formed  at  times  by  the  large  exhaust  pipe  between 
cylinders.  The  size  seems  to  have  no  effect  on  the  operation. 

DETERMINATION  OF  RELATIVE  SIZES  OF  CYLINDERS 

The  combined  card  without  clearance  for  a  triple  expansion 
engine  with  infinite  receivers  takes  the  form  shown  in  Fig.  115. 
The  whole  card  which  represents  the  entire  work  of  the  engine 
could  be   developed   alone  by 
the  low-pressure  cylinder.     The 
volume  of  this  cylinder  is  ab. 
The  volume  of  the  intermediate 
cylinder  is  cd  and  that  of  the 
high  is  ef  if  the  expansions  are 
complete  to  the  receiver  pres-    c 
sures.     If,    however,    the   vol-    a 
umes    were   cd'   and  ef  there 
would  be  the  free  expansion  gf 
in  the  high  and  hd'  in  the  inter- 
mediate.    In  many  cases  free 

expansion  is  used  as  it  reduces  the  sizes  of  the  cylinders  and 
therefore  the  sizes  of  the  parts.  There  is  not  a  great  loss  of  work 
area  due  to  this. 

The  receiver  pressures  are  fixed  by  the  points  of  cut-off  of  the 
lower  cylinders.  Thus,  with  a  given  cut-off,  in  the  high-pressure 
cylinder,  there  is  a  fixed  volume  of  steam  and  the  line  ifh  of 
the  expansion  of  this  steam  is  fixed.  If  now  in  any  cylinder 
the  volume  at  cut-off  is  known  this  must  be  on  the  line  of  ex- 


FIG.    115. — Intermediate   pressures 
with  infinite  receivers. 


248  HEAT  ENGINEERING 

pansion  and  consequently  the  pressure  is  determined.  If  the 
cut-off  is  made  later  in  the  intermediate  cylinder  the  point  / 
would  be  further  to  the  right  and  consequently  the  pressure 
would  have  to  be  lower.  This  would  produce  less  work  on  the 
cylinder.  Hence  the  statement  is  often  made  that  increasing 
the  cut-off  of  a  cylinder  reduces  the  work  of  that  cylinder.  If 
this  is  done  on  the  high-pressure  cylinder  the  same  result  follows, 
as  the  other  cut-offs  will  be  made  relatively  earlier  by  this  opera- 
tion and  the  intermediate  pressure  will  rise. 

The  whole  area  of  the  combined  card  being  the  work  which 
could  be  developed  by  the  low-pressure  cylinder  gives  a  method 
of  determining  the  size  of  the  cylinder.  The  size  of  this  cylinder 
fixes  the  power  of  the  engine.  The  sizes  of  the  other  cylinders 
determine  the  relative  amount  of  work  done  by  them  but  do 
not  affect  the  power  of  the  engine. 

The  m.e.p.  for  the  whole  card  is  found  by  assuming  pi,  pb,  and 
R,  the  total  ratio  of  expansion. 

.e.p.  =  /[|  (1  +  log,  R)  -pb]=P  (12) 

/  =  diagram  factor  X  clearance  factor. 
=  0.65  for  naval  triple. 
=  0.70  for  naval  compound. 
=  0.80  for  Corliss  compound  and  triple  land. 
=  0.90  for  pumping  engine. 

Another  form,  if  pr,  the  pressure  at  the  end  of  expansion  is 
assumed  rather  than  R,  is 

.e.p.  =  f[pr  (l  +  log,  g)  -pt]=P  (13) 

From  this  the  size  may  be  found  by  assuming  2LN  and  finding 

Ft. 

2PZJW 


m. 


m 


Having  Ft  the  areas  of  the  pistons  of  the  other  cylinders  are 
found  from  the  ratios  of 

ed'       ,  ef 


Thus  Fi  =  F'  (15) 


MULTIPLE  EXPANSION  ENGINES  249 

The  lengths  cd'  and  ef  are  found  by  obtaining  the  pressures 
7/2  and  p"z  and  drawing  the  figure  or  computing  the  lines  cd 
and  ef  and  allowing  for  the  distances  dd'  and  ff. 

Thus  cd  =  ab~  (17) 

Pr 

ef  =  ab  ^  (18) 

Pr 

There  are  four  ways  in  which  to  fix  the  intermediate  pressures  : 
(a)  equal  works,  (6)  equal  ratios  of  expansion,  (c)  equal  tempera- 
ture ranges  and  (d)  assumed  ratios. 

(a)  For  equal  work  the  card  is  divided  into  equal  areas. 

Area  card  =  D,[pr  (l  +  log,  ^)  -  p6]  =  Ft  (19) 

Area  1.  p.  card  =  Dt[pr  (l  +  loge  |^)  -  p6]  =  Ft  (20) 

Area  1.  p.  card  +  i.p.  card 

=  D,[pr  (l  +  log.  ^)  -  pt]  =  F2  (21) 
Area  of  three  lowest  cards 

=  D,  [pr  (l  +  log,  ^)  -  p»]  =  F3  (22) 

For  n  stages:  p,  =  -F,  (23) 

/Z' 

F2  -  ?F(  (24) 

F,  =  ?F,  (25) 

In  each  of  these  there  is  only  one  unknown  term  p2.  Hence 
these  pressures  may  be  found. 


Thus  i[pr  (l  +  loge  f)  -  pb]  =  pr  (l  +  log,  g  -  P6        (26) 


(28) 

In  the  same  manner: 

L-B  (29) 


250  HEAT  ENGINEERING 

For  the  mth  receiver: 

(30) 


(b)  For  equal  ratios  of  expansion  the  following  method  is  used  : 

7?  -  Pl 
K  ~  ^ 

PT 

R  =  ri  X  r2  X  r3  X  .....  =  rm 

r  =\/R  (31) 

p'z  =  rpr  (32) 

p"2  =  r*pr  =  rp'2  (33) 

p2«  =  /">  =>p2m~1  (34) 

(c)  For   equal   temperature   ranges   the   temperature   range 
between  pi  and  pb  is  found  and  this  divided  by  the  number  of 
stages  gives  the  range  per  stage.     If  this  is  added  to  the  lowest 
temperature  successively  the  various  temperatures  are  found 
and  from  these  the  corresponding  pressures  are  obtained. 

nr     —  T 

-  =  AT  (35) 

iiL 

T't  =  Tb  +  AT  (36) 

T"2  =  Tb  +  2A77  =  r2  +  AT  (37) 

Tm2  =  Tb  +  mAT  (38) 

(d)  For  assumed  ratios  of  cylinder  volumes  there  is  no  necessity 
of  finding  the  intermediate  pressure.     The  usual  ratios  of  practice 
are: 

Compound    engines     Dh  :  DI  =  1  :  2  to  1  :  7 
Triple  expansion    Dh  :  D{  ;  :  A  =  1  :  2.5  :  6.5 

For  compound  engines  a  ratio  1  :  4  is  quite  common. 

When  quick  response  is  needed  to  a  suddenly  changing  load 
the  small  ratios  are  used;  i.e.,  the  larger  high-pressure  cylinders 
are  used. 

As  an  example,  suppose  it  is  desired  to  construct  a  triple 
expansion  engine  to  produce  2000  kw.  from  a  generator  at  100 
r.p.m.  with  boiler  steam  at  175  Ibs.  gauge  pressure  with  80°  F. 
superheat,  with  the  pressure  at  release  —  5  Ibs.  gauge  and  a 
back  pressure  of  2  Ibs.  absolute. 


MULTIPLE  EXPANSION  ENGINES  251 

m.e.p.  =  0.90  [9.7(l  +  log,  ^-^  -  2]  =  33.0 

ihi      -  2000  X  1  6       33'°  X  100°  X  Fl 

33000 

Fi  =  3200  sq.  in. 

d  =  64  in. 
2LN  =  1000 

1000 

L  =  23000  =  5  ft'  =  60  in' 
Intermediate  pressures  and  volumes. 

Method  (a) 

logo's  =  (I  -  l)  [l  -  ~]  +  Ioge(l89.7  X  9.7*)" 

=  -  0.53  +  3.26 
=  2.73 
p'2  =  15.4. 

log,  P\  =   (|  -  l)  [g]  +  Ioge(l89.7  X  9.7W)M 

=  -  0.26  +  4.25 
=  3.99 
p\  =  52.1 

Fi  =  ^  1^4  =  320°  X  j^  =  2020;  cfc  =  51  in. 

Fh  =  3200  ^  =  595;  dh  =  28  in. 
Engine  28-51  and  64  X  60. 
Method  (6) 


r  =      l5  =  2.69 
pr2  =  2.69  X  9.7  =  26.1 
p\  =  (2.69)  2  X  9.7  =  70.2 
di  =  39  in. 
dh  =  24  in. 
Engine  24-39  and  64  X  60. 


252  HEAT  ENGINEERING 

Method  (c) 

T1  =  377.5  +  80  =  457.5 
!F2  =  126.2 


T'2  =  236.6;          p'2  =  23.5 
T"2  =  347.0;          p\  =  129.4. 
di  =  42  in. 
dh  =  17^  in. 
Engine  17^-42  and  64  X  60. 

Method  (d) 

Assume  ratio  of  1-2-6. 
di  =  45  in. 
dh  =  26  in. 
Engine  26-45  and  64  X  60. 

In  the  above  solutions  the  expansion  was  complete  in  the  upper 
two  cylinders.  If  the  diameters  are  made  slightly  smaller  there 
will  be  free  expansion  in  each  cylinder. 

The  relative  size  could  now  be  found  with  the  clearance  which 
varies  from  10  per  cent,  or  15  per  cent,  in  slide  valve  engines  to 
5  per  cent,  in  ordinary  Corliss  engines  and  2  per  cent,  in  engines 
with  the  valves  in  the  heads.  The  events  could  be  assumed  and 
the  volumes  of  the  receivers  after  which  the  cards  similar  to  those 
in  Fig.  113  could  be  computed. 

The  pressures  of  90  to  120  used  with  simple  expansion  engines 
are  changed  in  multiple  expansion  engines  to  from  150  to  225 
Ibs.  gauge.  The  higher  pressures  are  used  with  triple  expansion 
engines.  150  to  175  Ibs.  would  be  used  for  compound  engines 
although  higher  pressures  are  used  with  such.  At  times  lower 
pressures  are  used  with  these  two  stage  engines  but  generally 
lower  pressures  do  not  give  sufficient  expansion.  The  higher 
pressures  should  give  higher  efficiencies  but  the  troubles  from 
initial  condensation  and  free  exhaust  exist  here  to  a  great 
degree.  In  most  cases,  however,  neither  of  these  effects  is  so 
prominent  as  with  single  cylinder  engines  and  the  steam  consump- 
tion curves  are  more  nearly  flat,  giving  considerable  range 
without  much  variation  in  steam  consumption.  Such  steam 
consumption  curves  are  valuable  in  any  engine  or  turbine  as 
they  give  a  good  efficiency  over  a  great  range. 

The  back  pressure  should  be  reduced  to  as  low  a  value  as 


MULTIPLE  EXPANSION  ENGINES  253 

will  give  a  gain  on  the  entropy  diagram.  Owing  to  the  free 
expansion  in  these  engines  there  may  be  a  condition  for  which 
an  increase  of  vacuum  means  a  loss.  About  28  in.  is  sufficient 
for  the  vacuum  to  be  carried  on  a  multiple  expansion  engine. 

In  these  engines  as  in  the  case  of  the  simple  engines,  the  large, 
slow-speed  engines  are  those  which  give  the  highest  efficiency. 
The  larger  the  cylinder,  the  more  efficient  the  engine. 

Method  (a)  is  the  one  usually  employed  to  fix  the  relative 
size  of  cylinders. 

JACKETING 

In  these  engines  the  effect  of  jacketing  is  not  always  a  source 
of  economy  except  in  small  sizes.  The  gain  on  a  triple  expan- 
sion engine  is  probably  larger  than  that  found  on  a  compound 
engine  or  on  a  simple  engine  due  to  the  fact  that  heat  added 
to  the  exhaust  steam  by  the  jacket  when  the  moisture  on  the 
cylinder  walls  is  vaporized  is  of  use  in  the  lower  stages. 

A  number  of  tests  have  been  made  on  large  pumping  engines 
and  it  has  been  found  that  the  engine  would  consume  about  the 
same  amount  of  total  steam  with  or  without  jackets. 

REHEATERS 

Steam  as  it  leaves  the  cylinder  of  an  engine  is  likely  to  be  quite 
wet  and  hence  the  moisture  deposited  on  the  walls  at  entrance 
into  the  next  cylinder  would  be  increased  and  this  might  lead  to 
excessive  condensation.  To  cut  down  this  moisture,  the  receiver 
is  equipped  with  a  reheating  steam  coil  containing  high-pressure 
steam,  the  receiver  is  now  called  a  reheating  receiver  or  reheater. 
The  value  of  the  reheater  lies  in  the  fact  that  dry  steam  is  carried 
into  the  cylinder.  The  value  of  this  apparatus  is  also  questioned 
as  some  engines  on  test  show  an  increase  of  total  heat  when 
reheaters  are  used.  Other  engines  show  a  decrease  in  the  total 
heat.  The  heaters  should  contain  enough  heating  surface  and 
have  a  high  enough  steam  temperature  to  completely  dry  the 
steam  and  to  superheat  it  sufficiently  to  bring  the  steam  to 
a  dry  condition  at  cut-off.  If  this  can  be  done,  a  gain  should 
result. 

In  both  jackets  and  reheaters  the  apparatus  should  be  thor- 
oughly drained  and  the  supply  of  steam  should  be  taken  to  the 
apparatus  alone.  The  steam  should  be  hotter  than  the  steam  to 
be  heated.  The  use  of  steam  on  its  way  to  the  first  cylinder  for 


254 


HEAT  ENGINEERING 


a  jacket  is  bad  as  this  may  introduce  moisture  into  the  cylinder. 
The  use  of  exhaust  steam  for  a  jacket  is  condemned.  The  supply 
of  steam  should  be  through  pipes  of  ample  size.  This  apparatus 
should  always  be  air  vented.  Heavy  lagging  of  non-conducting 
material  has  been  found  of  value  in  engines  in  cutting  down 
radiation. 

GOVERNING 

The  multiple  expansion  engines  are  governed  by  changing 
the  point  of  cut-off  on  the  high-pressure  cylinder  alone  or  by 
changing  it  on  each  cylinder  at  the  same  time  and  in  the  same 
ratio. 

In  Fig.  116,  the  method  of  regulation  by  changing  the  cut-off 
on  the  high-pressure  cylinder  is  shown  while  the  method  by 


FIG.  116. — Method  of  governing  by 
changing  cut-off  on  high  pressure 
only. 


FIG.  117. — Method  of  governing 
by  changing  cut-off  proportionately 
on  each  cylinder. 


changing  the  cut-off  on  each  cylinder  is  shown  in  Fig.  117.  In 
this  latter  method  the  ratios  of  the  V's  are  all  equal  if  the  pressure 
in  the  receiver  is  to  remain  constant. 


In  the  first  method  the  receiver  pressure  changes  from  a  to  a' 
and  from  b  to  6'.  Before  the  receiver  pressure  can  drop  the  work 
done  must  be  represented  by  the  curves  shown  in  Fig.  118.  In 
this  it  will  be  seen  that  the  governor  has  reduced  the  area  by 
the  shaded  amount  and  if  this  is  enough  to  care  for  the  change  in 
load,  the  point  of  cut-off  will  have  to  gradually  change.  The 
receiver  pressures  gradually  change  to  their  final  values  and  the 


MULTIPLE  EXPANSION  ENGINES 


255 


expansion  curve  takes  the  position  shown  dotted  in  Fig.  118; 
the  area  between  the  original  and  final  curves  being  equal  to  the 
shaded  area.  Since  this  change  does  not  occur  at  once  with  re- 
ceivers of  any  size  the  action  is  sluggish.  Of  course  this  change 
is  accomplished  in  ten  or  fifteen  revolutions  but  it  will  not  be 
as  steady  as  that  resulting  from  the  method  shown  in  Fig.  117. 
With  small  fluctuations  either  method  is  good.  The  method  of 
Fig.  116  cuts  down  the  work  on  the  lower  cylinders  leaving  the 
high  pressure  about  the  same  while  the  method  of  Fig.  117 
affects  all  in  the  same  way  if  not  by  equal  amounts. 

Throttling  would  be  shown  in  Fig.  119.     In  this  the  pressure 
in  the  receiver  would  change  and  although  the  total  work  is 


FIG.  118. — Sluggish  action  due  to  cut- 
off change  on  high-pressure  cylinder  only. 
Shaded  area  shows  decrease  of  work  on 
first  stroke. 


FIG.    119. — Governing  by 
throttling. 


less,  the  reduction  is  mainly  on  the  low-pressure  cylinder,  as 
there  is  little  change  on  the  high.  This  method  throws  out 
the  equality  of  works. 

BLEEDING  ENGINES  OR  TURBINES 

The  use  of  low-pressure  steam  for  heating  water,  for  use  in 
warming  buildings  or  for  other  technical  applications  is  carried 
out  in  plants  having  high-pressure  steam,  by  throttling  the  steam 
through  a  reducing  pressure  valve  to  the  lower  pressure.  Now 
although  throttling  does  not  mean  a  loss  of  heat,  there  is  the 
loss  of  available  energy  if  by  this  is  meant  the  ability  to  be  turned 
into  mechanical  work.  This  may  be  seen  by  remembering  that 
it  is  a  difference  in  pressure  which  must  exist  to  give  ability  for 
work  in  a  cylinder.  The  production  of  high-pressure  steam  is 
usually  more  expensive  than  that  of  low  pressure  due  to  the 
higher  temperature  of  the  steam ;  hence  after  producing  this  high- 


256 


HEAT  ENGINEERING 


pressure  steam  many  engineers  believe  that  it  should  be  brought 
to  its  low  temperature  by  passing  it  through  an  engine  or  turbine 
where  its  available  energy  may  be  utilized.  After  reaching  the 
desired  temperature  and  pressure  this  steam  may  be  used.  If 
the  engine  or  turbine  is  connected  to  a 
condenser,  provision  is  made  to  take  off 
a  portion  of  the  steam  between  stages 
where  the  pressure  is  at  or  near  the 
atmospheric  pressure.  Were  this  amount 
constant  the  machines  could  be  made 
to  be  operated  with  one  quantity  to  one 
pressure  and  with  another  quantity  to  a 
lower  pressure.  The  card  from  such  an 

application    on  a   compound   engine  is 
FIG.  120 . — Combined      ,  •      -n«       i  <-»n       mi  i  i 

cards  from   a   compound    shown  in  Fig.  120.     The  volume  a  has 

engine  with  steam  removed  been  taken  for  use  outside  of  the  engine, 
from  receiver.  Cylinders  mv  i  v  i  i  T_ 

of  equal  sizes  -*-ne  low-pressure  cylinder  has   been  as- 

sumed to  be  similar  to  the  high-pressure 

cylinder  in  this  case.  If  no  steam  is  taken  the  low-pressure 
card  becomes  rectangular  as  shown  by  dotted  line. 

REGENERATIVE  ENGINES 

The  preceding  suggests  the  regenerative  steam  engine  cycle 
in  which  steam  is  taken  from  the  receiver  to  raise  the  feed  water 
to    that   temperature   before 
entering    the    boiler.     Steam 
from  one  receiver  brings  the 
water  to  that  temperature  and 
that   from   the  next  receiver 
brings    the    water    and    the 
steam  condensed  from  its  tem- 
perature to  the  temperature  of 
the  next.     To  see  the  applica- 
tion of  this  suppose  that  it  is 
applied  to  a  triple  expansion    FlG   121. —Combined  cards  from  a  re- 
engine   the   diagram  of  which        generative  triple  expansion  engine, 
is   shown    in  Fig.   121.     The 

decrease  in  the  volume  of  each  card  is  due  to  the  removal  of 
a  certain  amount  of  steam  to  mix  with  the  feed  water  to  raise 
it  to  the  temperature  of  the  receiver.  There  would  be  two 
analyses  of  this  depending  on  whether  or  not  the  condensed 


MULTIPLE  EXPANSION  ENGINES  257 

steam  from  the  feed  heaters  is  pumped  into  the  feed.  Of  course 
this  should  be  done.  The  expansion  is  assumed  adiabatic  in 
theory  and  1  Ib.  of  steam  will  be  assumed  on  the  lower  card. 
The  pressures  are  pi,  p2,  ps,  p^  and  p0. 

Heat  to  raise  1  Ib.  of  water  from  t0  to  fa  =  q'3  —  q'0 

Amount  of  steam  at  quality  x3  to  give  this  =  —  —  —  =  M2 

* 


Amount  of  steam  on  second  card  =  1  +  M2 

Amount  of  steam  of  quality  x2  to  raise  1  +  M2  Ibs.  of  water 


Amount  of  steam  on  first  card  =  1  +  M  2  +  MI 

Heat  to   raise   1  +  M2  +  MI  Ibs.  of  water  from  £2  to  t\  = 

(1  +  M2  +  AfOte'i  -q'*) 

Heat  to  vaporize  (1  +  M2  +  MI)  Ibs.  =  (1  +  Ml  +  M2)  xtfi 
Total  heat  from  outside  =  (1  +  M2  +  Mi)[ii  —  q'2]  =  Qi 
Work  done  on  cycles  =  (1  +  M2  +  M"i)b\  -  i2]  +  (1  +  M2) 

[ii  -  is]  +  l{[is  -  ij  +  A(p4  -  p0)v,}  =  AW 

The  i's  are  found  on  the  same  adiabatic  for  1  Ib.  of  steam  from 

one  end  to  the  other. 

Efficiency  =  ~^r~ 
Vi 

This  is  compared  with  efficiency  for  the  same  weight  in  each 
cylinder. 


Efficiency  =  (39) 

i\     q  o 

and  the  saving  is  shown. 

Let  this  be  applied  to  a  compound  engine  working  between 
150  Ibs.  absolute  and  2  Ibs.  absolute  with  original  dry  steam 
and  a  receiver  pressure  of  45  Ibs.  absolute  and  a  pressure  at  re- 
lease in  the  low  of  8  Ibs.  absolute. 

11  =  1192.6  qf!  =  330.0 

12  =  1100.0  q'2  =  243.7 
it  =  987.0  gr'j  =  150.9 
vs  =  39.9  q't  =    94.2 

x2  =        0.92  r2    =  927.5      . 

,,        243.7  -  94.2 
M2  =  0^2X-927^  =  °-176 

17 


258  HEAT  ENGINEERING 

Heat    =  1.176[1192.6  -  243.7]  =  1113 

Work  =  1.176[1192.6  -  1100]  +  ifllOO  -  987  +  ~~ 

I  i  O 

(8  -  2)39.9} 

=  108.7  +  157.4  =  266.1 

Theoretical  eff.  =  -7^  =  0.236 

144 


Theoretical  eff.  of  ordinary  cycle  = 


[1192.6  -  987  +  ==3(6  X  39.9)] 


778' 


250.0 
1098.4 


1192.6-94.2 


0.228 


LP.JHb.Bp, 
60 


FIG.   122. — Individual  and  combined  cards  from  test  of  pumping  engine. 

This  means  a  saving  of  0.8  in  22.8  or  3^  per  cent.  It  will  be 
seen  from  the  expression  for  the  work  on  each  cycle  that  the  low- 
pressure  cylinder  is  doing  50  per  cent,  more  work  than  the  high- 
pressure  cylinder. 

Data  from  the  test  of  a  triple  expansion  engine  with  its  cards 
given  in  Fig.  122; 


MULTIPLE  EXPANSION  ENGINES  259 

Size  of  pump 20,000,000  gal.  per  24  hr. 

Size  of  cylinders. 
'  Steam 30  in.,  60  in.  90  in.,  X  66  in: 

Water 33  in.  X  66  in. 

Pressures  by  gauge 

Steam  supply 180 . 2    Ibs.  per  sq.  in. 

1st  Receiver 25.4    Ibs.  per  sq.  in. 

2d  Receiver 9.5    in.  vacuum 

Condenser 26 . 7    in.  vacuum 

Water  discharge 86 . 9    Ibs.  per  sq.  in. 

Barometer 14. 86  Ibs.  per  sq.  in. 

Revolution  in  24  hr 28,989 

Displacement  per  revolution 732  gal. 

Temperatures. 

Return  from  condenser 113°  F. 

Water  pumped 40°  F. 

Condensing  water 40°  F. 

Water  leaving  condenser 70°  F. 

Outside  air 30°  F. 

Quality  of  steam 0 . 989 

Water  pumped  to  boiler,  per  hour 9,266  Ibs. 

Boiler  leakage,  per  hour 196  Ibs. 

Drip  in  engine  room,  per  hour 135  Ibs. 

Steam  to  drip  pump,  per  hour 58  Ibs. 

Net  boiler  steam,  per  hour 8,877  Ibs. 

Jacket  drip,  per  hour 661  Ibs. 

1st  Receiver  drip,  per  hour 422  Ibs. 

2d  Receiver  drip,  per  hour 361  Ibs. 

Indicated  horse-power  steam  end 

High  pressure 332 . 02 

Intermediate 269 . 10 

Low  pressure 260. 22 


Total 861.34 

Water  end 839 . 8 

Delivered  horse-power 822 . 9 

Steam  per  i.h.p.-hr 10. 37  Ibs. 

Heat  per  i.h.p.-min 193 . 04  B.t.u. 

Duty  per  1000  Ibs.  dry  steam 181,000,000 

TESTING  AND  ANALYSIS 

There  are  a  few  points  which  must  be  brought  out  in  regard  to 
testing  multiple  expansion  engines. 

The  exhaust  from  these  engines  is  not  all  at  the  temperature  of 


260  HEAT  ENGINEERING 

the  condensed  steam  when  reheaters  or  jackets  are  used.  In 
such  a  case  the  drips  from  the  jacket  or  reheater  are  at  tempera- 
tures much  higher.  These  possess  the  temperature  correspond- 
ing to  the  steam  pressure.  Hence  if  M  pounds  of  steam  are  used 
by  the  main  engine,  and  m/  and  mr  are  the  amounts  used  in 
the  jackets  and  reheaters,  the  amount  of  heat  chargeable  to  the 
engine  is 

Qi  =  (M  +  my  +  mr)ii  —  Mq0  —  irijq'j  —  mrqfr        (40) 


In  this  q'j  and  q'r  are  the  heats  of  the  liquid  in  the  jacket  and 
reheater. 

The  efficiency  which  has  been  given  as 

_  _  AW 

77  ~  (M  +  my  +  rnr}(i  -  qf0) 
AW 
is  n°W       *  =:    (M  +  my  +  mr)  i  -  Mq'0  -  mrf,  -  mrqfr 


In  Hirn's  analysis  for  multiple  expansion  engines  a  new  device 
must  be  used  to  find  the  heat  delivered  to  the  second  cylinder 
and  discharged  from  the  first  cylinder.  In  analyses  of  multiple 
expansion  engines  it  is  necessary  to  find  Qr,  the  heat  radiated  from 
each  cylinder.  Having  this  for  the  first  cylinder,  the  heat  losses, 
Qa,  Qb,  and  Qd,  are  found  by  the  formulae  and  then  Qc  is  given  by 

Qc  =  Qr  -  (Qa  +  Qb  +  Qd)  (43) 

The  heat  exhausted  from  the  first  cylinder  is 

-  AU3  +  AWC  -  Qc  (44) 


or  Q2  =  Q!  -  Qr  -  A(Wa  +  Wb  +  Wc  +  Wd)  (45) 

The  heat  given  to  the  second  cylinder  Qf,  is 

Q'l  =  Q2  +  Qh  -  Qi  (46) 

Qh  is  the  heat  given  to  the  steam  in  the  reheater  and  Qi  is 
the  heat  lost  in  radiation  from  the  reheater.  This  term  Qi  is 
found  by  a  radiation  test  similar  to  that  made  on  the  engine. 

By  observing  the  condensation  from  the  receiver  coil  when  no 
steam  is  passing  from  one  cylinder  to  the  other,  the  value  of 
Qi  is  found  while  the  observations  during  the  operation  of  the 
engine  gives  the  term  Qh. 

Qh  or  Qr  =  M(i  -  q'0)  (47) 


MULTIPLE  EXPANSION  ENGINES  261 

Equation  (45)  furnishes  the  heat  supplied  the  second  stage  and 
a  similar  procedure  is  used  for  the  lower  stages. 

If  a  jacket  is  used  on  any  cylinder,  the  heat  supplied  by  the 
jacket  is  found  by  weighing  the  steam  condensed.  The  heat  is 
given  by 

Qi  =  M(i-q')  (48) 

If  an  engine  is  jacketed  there  is  a  new  term  in  Qr: 

Qr    =    Qa+Qb  +   Qc+Qd   +   Q3  (49) 

and  for  multiple  expansion  engines  Qc  is  given  by 

Qc  =  Qr  -  (Qa  +  Qb  +  Qd  +  Q,-)  (50) 

Then     Q2  =  Qi  +  Qi  -  Qr  -  A(Wa  +  Wb  +  Wc  +  Wd)       (51) 

BINARY  ENGINES 

Another  method  of  utilizing  the  waste  heat  and  avoiding  the 
waste  due  to  free  expansion  on  account  of  the  great  volume  of 
low-pressure  steam  is  to  use  the  exhaust  to  volatilize  a  liquid, 
such  as  sulphur  dioxide,  which  has  a  much  higher  saturation 
pressure  than  steam.  In  this  way  the  same  limits  of  temperature 
may  be  utilized  without  passing  to  such  low  pressures  as  used  on 
steam  engines.  Engines  using  two  vapors  are  known  as  Binary 
Engines.  In  a  development  of  this  engine  by  Professor  E.  Josse 
of  Berlin,  steam  from  an  engine  was  exhausted  into  a  surface  con- 
denser in  which  the  condensing  fluid  was  volatile  S02,  which  by  its 
evaporation  removed  heat  from  the  steam  and  condensed  it.  The 
pressure  of  the  steam  was  3  Ibs.  per  square  inch.  This  gave  a  tem- 
perature of  141°  F.  SO2  at  132°  F.  is  under  pressure  of  132  Ibs. 
per  square  inch  so  that  the  S02  would  boil  and  produce  a  pressure 
of  132  Ibs.  absolute  while  condensing  steam  at  3  Ibs.  pressure. 
This  could  be  used  in  a  cylinder  connected  with  the  same  shaft  as 
the  steam  cylinders  and  aid  in  the  driving.  This  cylinder  ex- 
hausted into  a  condenser  cooled  by  water  at  50°  F.  in  which 
the  SO2  condensed  at  a  pressure  of  31  Ibs.  absolute  and  a  tempera- 
ture of  66°  F.  The  condensed  SO2  was  pumped  into  the  steam  con- 
denser where  it  was  evaporated  again  and  used. 

In  the  steam  engine  Josse  developed  about  150  h.p.  on  250 
B.t.u.  per  horse-power  minute  and  by  using  the  exhaust  as  de- 
scribed above,  50  additional  horse-power  were  developed  giving 
176  B.t.u.  per  horse-power  minute. 


262  HEAT  ENGINEERING 

The  reason  for  the  'gain  in  the  binary  engine  is  found  in  the 
fact  that  the  effect  of  initial  condensation  on  a  large  cylinder 
or  the  effect  of  free  expansion  have  been  eliminated.  These 
are  practical  reasons  and  not  theoretical.  The  range  of  tempera- 
ture has  not  been  increased  but  the  pressure  and  volume  limits 
have  been  so  changed  that  these  engines  have  a  high  practical  effi- 
ciency, 171.  ' 

Josse  has  installed  the  engines  commercially  and  has  found  that 
they  are  valuable  provided  a  high  load  factor  can  be  had.  For 
an  intermittent  load  or  for  use  at  intervals  the  engine  is  too 
expensive. 

TOPICS 

Topic  1.  —  What  are  multiple  expansion  engines?  Why  were  they  used? 
Why  are  they  of  value?  Explain  how  the  combined  diagram  is  constructed. 
What  is  the  difference  between  the  Woolf  and  the  receiver  compound 
engines? 

Topic  2.  —  Sketch  the  combined  cards  for  a  Woolf  compound  engine.  Write 
the  formulae  by  which  the  pressures  at  the  various  points  may  be  computed. 
Explain  why 

paVa  +  p,V,  _ 
va+vv        ~pr' 

Topic  3.  —  Sketch  the  combined  cards  for  a  receiver  compound  engine. 
Write  the  formulae  by  which  the  pressures  at  the  various  points  may  be  com- 
puted. Explain  why 

y    =  p 


Va+Vv 

Topic  4.  —  Explain  why  it  may  be  said  that  the  size  of  the  low-pressure 
cylinder  fixes  the  power  of  a  multiple  expansion  engine.  On  what  does  the 
size  of  the  higher  pressure  cylinders  depend?  If  the  size  of  the  high-pressure 
cylinder  is  changed  what  is  the  effect  of  this? 

Topic  6.  —  Derive  the  formulae  for  the  total  work  of  an  n-stage  engine  and 
for  the  work  on  the  low-pressure  cylinder.  From  these  find  the  pressure  at 
entrance  to  the  low-pressure  cylinder. 

Topic  6.  —  Give  the  method  of  finding  the  intermediate  pressures  for  equal 
ratios  of  expansion,  equal  temperature  ranges  and  given  ratios  of  volume. 
Knowing  the  pressures,  show  how  the  volumes  of  the  various  cylinders  are 
found  assuming  (a)  complete  expansion  and  (6)  some  free  expansion.  What 
is  the  reason  for  assuming  receivers  of  infinite  capacity? 

Topic  7.  —  Why  is  a  28-in.  vacuum  a  limit  for  vacuum  in  steam-engine 
operation?  What  is  the  effect  of  jackets?  Reheating?  Explain  what 
happens  when  the  cut-off  is  made  later  on  any  cylinder  and  the  reason  for 
this  effect. 

Topic  8.  —  Discuss  with  diagrams  the  various  methods  of  governing  multi- 
ple expansion  engines  and  show  which  is  the  better  form. 


MULTIPLE  EXPANSION  ENGINES  263 

Topic  9. — Explain  what  is  meant  by  bleeding  engines  or  turbines  and  why 
this  is  of  value.  Explain  the  action  of  the  regenerative  engine.  Give  the 
expressions  for  the  various  amounts  of  heat  and  work  and  derive  the  expres- 
sion for  efficiency. 

Topic  10. — Derive  the  expression  for  the  heat  delivered  to  one  of  the  lower 
stages  of  a  multiple  expansion  engine  (Q\  or  Qzoi  Hirn's  analysis).  Show 
what  effects  jackets  or  reheaters  have  on  this  formula.  What  heat  is  charge- 
able to  an  engine  when  the  drips  from  the  jacket  and  receivers  may  be  sent 
back  to  the  feed  ?  What  is  a  binary  engine  ? 

PROBLEMS 

Problem  1. — Find  the  pressures  at  the  corners  and  events  of  the  stroke 
for  a  compound  engine,  8  in.  and  18  in.  X  24  in. — 80  r.p.m.,  with  a  re- 
ceiver of  volume  equal  to  twice  the  volume  of  the  high-pressure  cylinder 
and  with  8  per  cent,  clearance  on  the  high-pressure  cylinder  and  4  per  cent, 
on  the  low.  Initial  gauge  pressure  is  125  Ibs.  and  the  back  pressure  is  2 
Ibs.  absolute.  Cut-off  is  25  per  cent,  and  compression  is  at  20  per  cent,  of 
the  stroke  from  the  end.  Cranks  at  right  angles. 

Problem  2. — Construct  the  combined  cards  for  Problem  1  to  scales  of  1  in. 
=  30  Ibs.  and  1  in.  =  0.435  cu.  ft.  Find  the  m.e.p.  of  each  card.  Construct 
the  h.p.  card  to  scales  of  1  in.  =  60  Ibs.  and  1  in.  =  0.1745  cu.  ft.  and  the 
l.p.  card  to  scales  of  1  in.  =  20  Ibs.  and  1  in.  =  0.884  cu.  ft.  Find  the 
scale  of  B.t.u.  per  square  inch  of  area  of  card,  and  of  ft. -Ibs.  per  square  inch 
of  card.  Find  the  h.p.  of  the  engine. 

Problem  3. — Find  the  equivalent  m.e.p.  to  be  expected  in  Problem  1  if 
the  complete  ratio  of  expansion  is  10.  What  is  the  size  of  the  low-pressure 
cylinder  to  develop  350  i.h.p.? 

Problem  4. — Find  the  size  of  the  low-pressure  cylinder  of  a  compound 
Corliss  engine  to  drive  a  2000-kw.  generator.  The  initial  pressure  is  160 
Ibs.  gauge,  the  back  pressure  is  3  Ibs.  absolute,  and  the  pressure  at  end  of 
expansion  in  the  low-pressure  cylinder  is  —4  Ibs.  gauge.  Find  the  size  of  the 
high-pressure  cylinder  by  four  methods. 

Problem  6. — Find  the  receiver  pressures  for  a  triple  expansion  engine 
operating  between  175  Ibs.  gauge  and  2  Ibs.  absolute,  with  a  complete  ratio 
of  expansion  of  22. 

Problem  6. — Construct  a  p-v  diagram  for  Problem  5  with  infinite  receiver 
and  no  clearance  and  compression.  Construct  the  T-S  diagram  for  the  same 
cards  and  show  whether  or  not  it  would  pay  to  reduce  the  back  pressure  to 
1  Ib.  absolute. 

Problem  7. — Find  the  sizes  of  cylinders  to  be  used  with  a  compound  engine 
developing  1000  h.p.  if  the  steam  consumption  of  the  engine  is  15  Ibs.  per 
1  h.p.-hr.  when  30  per  cent,  of  the  steam  is  removed  from  the  receiver  and 
the  engines  develop  equal  works.  The  limits  of  pressure  are  150  Ibs.  gauge 
and  3  Ibs.  absolute. 

Problem  8. — Find  the  thermal  efficiency  of  a  triple  expansion  regenerative 
engine  with  gauge  pressures  as  follows:  Initial  high,  170  Ibs.;  first  receiver, 
65  Ibs. ;  second  receiver,  5  Ibs. ;  end  of  expansion,  —  8  Ibs. ;  back  pressure, 
-  12  Ibs. 


264  HEAT  ENGINEERING 

Problem  9. — Design  a  cylinder  to  deliver  120  i.h.p.  at  100  r.p.m.  with 
pi  =  130  Ibs.  gauge,  pv  —  3  Ibs.  gauge,  r  =  5.  Find  the  probable  x  at  cut- 
off. Find  the  probable  steam  consumption.  Find  the  pressure  of  SO2 
evaporated  by  the  heat  of  the  exhaust.  Find  the  back  pressure  in  an  engine 
working  with  this  S02  if  the  condensing  water  is  70°  F.  Sketch  a  card  for 
the  SO  2  cylinder.  Find  the  volume  of  SO2  vapor  per  minute  at  release  and 
from  this  find  the  horse-power  of  the  SO2  cylinder  and  its  size  at  100  r.p.m. 
What  is  the  thermal  efficiency  of  the  combined  engine?  At  for  heat  trans- 
mission is  15°  F. 


CHAPTER  VII 

STEAM  NOZZLES,  INJECTORS,  STEAM  TURBINES 

NOZZLES 

In  Chapter  I  the  formula  for  the  velocity  of  discharge  of  any 
fluid  through  a  passage  from  pressure  p\  to  pressure  p%  was  de- 
rived. This  gave 

(i) 


ii  and  iz  are  the  heat  contents  at  the  two  pressures  regardless 
of  whether  or  not  there  is  internal  friction  but  dependent  only  on 
the  fact  that  there  is  no  external  heat  transfer.  It  was  pointed 
out  that  iz  for  any  pressure  was  difficult  to  find  if  there  was  in- 
ternal friction  and  hence  the  ordinary  method  was  to  find  the 
value  of  iz  on  a  reversible  adiabatic  (which  means  no  friction) 
through  pi  ii,  and  to  subtract  from  the  heat  (ii  —  iz),  which 
should  have  been  transformed  into  kinetic  energy,  the  amount 
utilized  in  friction.  This  gives  the  amount  remaining  for  change 
in  kinetic  energy.  If  y(ii  —  iz)  is  the  amount  of  energy  used  in 
internal  friction  against  the  sides  of  the  passage  and  between  the 
particles  of  the  fluid,  the  amount  left  for  the  change  of  kinetic 
energy  is 

(ii  -  12)  (1  -  y)  (2) 

This  quantity  represents  the  change  in  kinetic  energy  and  gives 
w*  =  223.7-v/fo  -  12)  (1  -  y)  (3) 

Wi  ^ 

when  i0i  is  so  small  that  ^—  may  be  neglected  or 


I  Aw  2 

.7-^  fe  -  i2)(l  -  y)  +  ~- 


wz  =  223. 

when  wi  is  appreciable.     If  Wi  is  200  ft.  per  second  it  is  inap- 
preciable as     ^  1    is  0.8  while  the  first  term  of  the  bracket  may 

be  from  50  to  100  or  even  250  B.t.u.     The  importance  of  this 
term  is  determined  by  the  conditions  of  any  problem. 

265 


266 


HEAT  ENGINEERING 


The  value  of  y  is  determined  by  experiment.  w2  is  measured 
by  the  reaction  of  the  jet  and  weight  of  the  steam  or  by  a  pitot 
tube  and  from  this  measured  value  of  w<z,  the  value  of  y  is  com- 
puted. By  measuring  the  quality  of  the  steam  actually  present 
at  2  and  comparing  it  with  that  of  adiabatic  expansion  the  value 
of  y  may  be  found.  In  most  cases  y  will  vary  from  0.10  to  0.15 
for  long  nozzles  with  large  angles  of  divergence  or  for  length  of 
10  diameters  for  small  angles  of  divergence  to  0.06  or  0.08  for 
shorter  tubes.  It  seems  that  with  nozzles  the  coefficient  de- 
creases with  the  increase  of  pressure.  With  orifices  in  plates  the 
value  of  this  y  may  be  15  per  cent. 

Suppose  that  the  pressure  drop  is  uniform  along  a  tube  and  that 
frictional  loss  is  zero.  If  the  velocity  at  any  point  be  computed 
by 

wx  =  223.7\/(«i  -  ix)  (5) 

and    1    Ib.    of    steam    is   assumed   to    be   passing    per   second 
the  area  may  be  computed  by 

_  MVX  _  Mxv"x  _  irdx2 
x         wx  wx  4 


(6) 


The  values  at  different  points  for  the  above  assumptions  are 
tabulated  below. 

NOZZLE  DATA 


Pressure 
absolute 

Quality 

Heat 
content 

Specific 
volume 

Velocity 

Area, 
sq.  in. 

Diam., 
meters 

149    1 

0  4°  sup 

1193   3 

3   04 

124.4 

0.987 

1178.3 

3.55 

865 

0.592 

0.87 

100.2 

0.972 

1161.0 

4.30 

1270 

0.487 

0.79 

75.4 

0.953 

1138.8 

5.51 

1640 

0.484 

0.78 

50.0 

0.929 

1108.1 

7.91 

2060 

0.554 

0.84 

24.97 

0.894 

1059.3 

14.58 

2590 

0.810 

1.02 

14.99 

0.872 

1026.0 

22.91 

2890 

1.140 

1.20 

4.97 

0.830 

959.9 

61.20 

3410 

2.580 

1.81 

The  diagram  of  Fig.  123  will  be  obtained,  in  which  velocity, 
specific  volume,  area,  and  radius  are  plotted.  It  is  to  be  re- 
membered that  in  this  figure  the  pressure  drop  has  been  made 
uniform.  The  values  of  ii  and  iz  were  found  in  the  same  entropy 
column  and  xxv"x  was  found  in  that  column  or  computed  from 


STEAM  NOZZLES,  INJECTORS,  STEAM  TURBINES       267 


the  value  of  x.  It  will  be  seen  that  the  required  area  reaches  a 
minimum  section  and  then  increases.  This  is  due  to  the  fact  that 
the  specific  volume  curve  does  not  increase  as  rapidly  as  the  ve- 
locity until  the  critical  point  of  discharge  is  reached  and  after 
this  the  more  rapid  increase  of  specific  volume  makes  it  necessary 


150 

I     75 
0 

3000 

12000 

1000 

I  15'° 
*o 

t  10.0 
I    5.0 

1.5 

I" 

0.5 
0 

*  0.75 
1 

0.25 
0 


1  0 


90 


Radii 


of  N 


GO 


zzle 


30 


Length 

FIG.  123. — Curves  of  velocity,  volume,  area  and  diameter  for  nozzle 
assuming  uniform  pressure  drop. 

to  increase  the  area.     This  critical  point  is  at  the  pressure  point 
2 


Pt 


when 


/     2     \    n 
=  ( — TTTJ71"1  PI  =  0.57pi  for  steam 

n  =  1.135. 


(7) 


Whenever  pz  is  less  than  this  critical  value  there  must  be  a 
minimum  section  in  the  tube.     Such  a  tube  is  spoken  of  as  a 


268 


HEAT  ENGINEERING 


nozzle  and  the  small  section  is  known  as  the  throat  and  the  large 
end  is  known  as  the  mouth.  On  account  of  the  high  pressure 
available  at  the  entrance  to  a  nozzle  the  section  from  the  entrance 
to  the  throat  is  made  very  short.  The  length  of  this  is  usually 
made  about  equal  to  one  or  two  times  the  diameter  of  the  throat. 
A  nozzle  may  be  assumed  of  the  section  shown  in  Fig.  124,  and 
the  areas  at  various  points  are  found  and  plotted.  At  various 
assumed  pressures  the  velocity  is  determined  and  then  the  area 
to  carry  1  Ib.  of  steam  is  computed  by  the  formulae: 


w2  =  223.7- 


(1) 

(6) 


The  position  on  the  curve  of  area  for  this  value  will  fix  the  posi- 
tion of  the  assumed  pressure.  Following  this  for  the  various 
points  the  curve  of  pressure  is  obtained.  The  following  table 
gives  the  computation  for  the  curve,  using  Peabody's  Entropy 
Table  for  entropy  1.56  and  running  from  160.8  to  14.7  Ibs. 
pressure. 


Pres- 

160.8 

121.1 

90.9 

50.0 

30.35 

20.02 

14.7 

10.17 

sure 

ii 

1191.2 

1191.2 

1191.2 

1191.2 

1191.2 

1191.2 

1191.2 

1191.2 

12 

1191.2 

1168.1 

1145.4 

1100.7 

1065.5 

1037.7 

1018.0 

995.2 

ii  —  i* 

0 

23.1 

45.8 

90.5 

125.7 

153.5 

173.2 

196.0 

wz 

0 

1071.0 

1510.0 

2120.0 

2500.0 

2780.0 

2940.0 

3130.0 

X2V"2 

2.811 

3.601 

4.63 

7.84 

12.18 

17.60 

23.13 

32.08 

Area 

00 

3.36 

3.06 

3.70 

4.87 

6.12 

7.87 

10.2 

-MO3 

At  0.57  of  160.8  or  91.7  Ibs.  the  area  is  0.00306  sq.  ft. 
Napier's  formula  this  area  would  have  been 


By 


F  = 


70X1 


160.8  X  144 


a  result  practically  the  same  as  that  in  the  table  above.  In 
drawing  the  nozzle  of  Fig.  124  the  area  at  the  end  has  been  made 
0.00985  sq.  ft.  which  is  larger  than  0.00787  sq.  ft.  required  at  14.7 
Ibs.  pressure.  This  is  known  as  overexpansion  and  the  steam 
will  expand  to  below  the  pressure  at  the  end  and  be  brought  up 
again  as  shown  in  the  figure.  It  means  a  loss  in  efficiency.  This 
effect  is  greater  than  the  loss  due  to  underexpansion  as  will  be 


STEAM  NOZZLES,  INJECTORS,  STEAM  TURBINES       269 


seen  later.  The  reverse  curve  near  the  throat  indicates  that  the 
distance  from  entrance  to  throat  or  minimum  section  is  too  long. 
A  shorter  length  giving  the  steeper  curve  would  be  better. 

If  now  friction  be  assumed  and  the  areas  be  found  for  different 
pressures  by  the  formulae: 

ft  =  ^Tp  (6) 


the  curve  of  pressure  drop  with  friction  may  be  had  for  any  given 
form  of  nozzle  in  the  same  manner  as  that  used  above.  In 
formula  (6)  the  value  of  zV '2  is  not  found  in  the  same  entropy 


o.ooio 


0.005 


150 


100 


50 


zzle  Section 


FIG.  124.— Pressure  curve  of  nozzle  obtained  from  velocities  and  areas. 

column  or  line  as  that  on  which  ii  was  found  because  in  this  case 
the  heat  content  at  discharge  is  i*  of  the  isoentropic  line  for  the 
pressure  p2  plus  the  friction  y(ii  —  iz).  If  curves  for  various 
values  of  y  are  computed  and  these  curves  are  compared  with  the 
actual  pressure  at  various  points,  the  probable  values  of  y  may 
be  found. 

Stodola  has  explored  the  different  points  of  a  nozzle  by  an 
exploring  tube  introduced  along  the  axis.  This  tube  was  closed 
at  the  inner  end  and  connected  to  a  pressure  gauge  at  the  outer 


270 


HEAT  ENGINEERING 


end.  Holes  were  bored  into  this  around  the  circumference  at 
one  point  in  the  length.  In  place  of  being  normal  to  the  tube  or 
at  right  angles  to  the  axis  they  were  inclined  at  45°.  There  were 
two  tubes,  one  with  the  holes  inclined  in  the  direction  of  flow 


FIG.  125. — Stodola's  apparatus  for  exploring  nozzles. 

and  one  with  them  inclined  against  the  direction  of  flow.  The 
readings  of  these  varied,  due  to  the  impact  of  the  steam  increasing 
the  pressure  in  one  and  decreasing  it  in  the  other.  The  mean 
value  was  used.  In  addition  normal  holes  were  introduced  into 


STEAM  NOZZLES,  INJECTORS,  STEAM  TURBINES       271 


the  side  wall  of  one  tube  as  shown.  The  pressures  at  the  large 
end  of  the  nozzle  showed  that  for  this  part  the  curve  of  the  value 
of  y  =  0.10  agreed  closely  with  the  actual  curve  although  there 
was  a  slight  increase  in  y  as  the  end  was  reached. 

Stodola  found  in  this  experiment  that  the  pressures  at  the  wall 
of  the  nozzle  were  practically  the  same  as  those  at  the  center  of  the 
nozzle  showing  that  with  the  conical  nozzle  there  was  pressure 
exerted  on  the  wall  confining  the  steam.  If  there  were  no  wall 
present  as  in  the  case  of  discharge  from  a  hole  in  a  plate  the 
steam  would  be  forced  outward  and  would  follow  a  curved  path, 


110  Ibs 


150 Ibs. 


100  Ibs. 


50  Ibs. 


100  Ibs. 


50 Ibs. 


FIG.  126. — Stodola's  exploration 
curves  for  nozzles. 


FIG.  127.— Stodola's  exploration 
curves  for  orifices. 


the  centrifugal  force  exerting  the  necessary  radial  pressure  to 
allow  the  steam  to  have  an  acceleration  axially.  This  cen- 
trifugal force  is  only  another  way  of  expressing  the  inertia  of  the 
steam  against  radial  acceleration. 

If  a  valve  is  put  in  the  pipe  line  beyond  the  nozzle,  the  pres- 
sure of  the  discharge  region  may  be  raised,  and  the  curve  found 
by  the  exploring  tube  shows  what  may  happen  in  over  expan- 
sion. Curves  obtained  by  Stodola  are  shown  in  Fig.  126. 

The  waves  set  up  at  low  pressures  are  due  probably  to  acous- 
tic vibrations  which  are  so  dampened  by  friction  as  to  be  elimi- 
nated in  a  short  time.  This  is  due  to  the  fact  that  the  velocity  at 


272 


HEAT  ENGINEERING 


the  critical  pressure  of  the  throat  corresponds  to  the  velocity  of 
sound.  This  velocity  of  course  depends  on  the  pressure  and  tem- 
perature of  the  substance.  Stodola  points  out  that  this  is  an 
illustration  of  Biemann's  Theory  of  Steam  Shock. 

If  an  investigation  by  the  exploring  tube  is  made  on  a  hole 
with  a  rounded  entrance,  known  as  an  orifice,  shown  in  Fig.  127, 
curves  of  somewhat  similar  forms  are  found  and  it  will  be  noted 
that  the  pressure  in  the  plane  of  the  orifice  is  practically  constant 
although  it  reaches  a  lower  value  beyond  the  face  of  the  orifice. 
With  a  sharp-edged  orifice  the  same  observations  were  made 
although  here  the  pressure  at  the  outer  plane  of  the  orifice  seemed 
to  be  the  critical  pressure.  The  crossing  of  the  pressure  lines  at 
the  mouth  of  the  orifice  means  that  the  velocity  at  this  point  is 
the  same  whatever  be  the  outer  pressure,  hence  the  quantity 
discharged  will  be  the  same  whatever  the  lower  pressure  is,  pro- 


FIG.  128. — Mouthpiece  or  orifice. 


FIG.  129. — Rateau's  orifices. 


vided  it  is  less  than  the  critical  pressure.  Above  this  the  line  of 
pressure  at  the  crossing  of  the  plane  of  the  orifice  is  greater  and 
varies  with  the  pressure  pz'}  the  velocity  in  the  tube  is  less  than 
before.  The  great  drop  in  pressure  in  this  figure  shows  that  the 
axial  distance  to  the  point  of  critical  pressure  may  be  very  short 
and  hence  the  arrangement  of  the  nozzle  in  Fig.  124  was  pos- 
sible. It  could  have  been  made  shorter.  The  existence  of  this 
throat  demands  a  greater  total  pressure  drop  to  the  end  than  the 
amount  to  bring  the  pressure  to  the  critical  point.  Hence  a 
nozzle  of  the  form  of  Fig.  124  would  mean  excessive  overexpan- 
sion  if  the  pressure  p^  were  not  below  the  critical  pressure,  as 
shown  in  Fig.  126  with  a  high  back  pressure.  This  would  mean 
shock  and  resultant  loss.  In  the  case  of  a  drop  in  pressure  to  a 
point  above  the  critical  pressure,  the  expansion  part  of  the 


STEAM  NOZZLES,  INJECTORS,  STEAM  TURBINES       273 

nozzle  is  omitted  giving  the  orifice  of  Fig.  128.  On  account  of  the 
pressure  not  reaching  the  critical  point  in  certain  cases  it  may  be 
well  to  stop  the  curves  at  points  earlier  than  the  section  with  the 
parallel  elements.  This  would  give  the  forms  suggested  by 
Rateau,  shown  in  Fig.  129.  These  are  known  as  orifices  or 
converging  nozzles. 

In  these,  the  converging  portion  reduces  the  steam  pressure  to 
a  point  above  the  critical  pressure.  Rateau  showed  that  the 
coefficient  of  discharge  of  these  nozzles  varied  from  0.94  for  a 
small  pressure  drop  to  unity  at  the  critical  point  where  p2  = 
0.57  pi.  For  the  thin  orifice  shown  in  the  figure  the  coefficient 
of  discharge  was  0.82  at  its  critical  point.  This  means  that  in 
designing  nozzles  the  first  step  is  to  find  the  relation  between 
the  pressure  pz  and  p\.  If  p2  is  greater  than  0.57pi,  the  nozzle 
is  of  the  form  shown  in  Fig.  129.  That  is,  it  is  an  orifice  or 
mouthpiece.  If  p%  is  less  than  0.57pi  there  will  be  a  minimum 
section  or  throat  and  there  is  a  diverging  part,  best  spoken  of  as 
a  nozzle. 

The  velocity  and  area  are  computed  for  the  throat  of  the  orifice 
or  the  throat  and  mouth  of  the  nozzle.  The  steps  will  be  best 
illustrated  by  two  problems. 

Suppose  a  nozzle  is  to  be  designed  to  discharge  10  Ibs.  of  steam 
per  second  from  160.8  Ibs.  absolute  pressure  and  quality 
0.9966  into  a  vacuum  of  20".  This  corresponds  to  4.91  Ibs. 
absolute.  The  areas  are  required. 

p2  <  0.57pi  or  91.66  Ibs. 

Hence  there  is  a  throat  as  well  as  a  mouth. 

In  figuring  the  velocity  at  the  throat,  the  length  of  this  is  so 
short  for  the  drop  in  pressure  that  the  friction  is  negligible  while 
for  the  main  length  of  the  diverging  portion  there  must  be  a  value 
of  y  as  mentioned  on  p.  266.     The  y  for  the  problem  considered 
with  a  long  tube  and  large  drop  is  0.15.     For  a  drop  of  a  little 
less  than  this  from  the  critical  pressure  the  value  of  y  would  be 
about  0.075  if  there  were  a  short  diverging  portion  to  the  nozzle. 
wt  =  223.7\/(1191.2  -  1146.0)  =  1501  ft.  per  sec. 
wm  =  223.7V  (1191.2  -  953.1)0.85   =  3175  ft.  per  sec. 

vt  =  4.60 

vm  is  found  on  the  line  for  4.91  Ibs.  pressure  but  at  a  value  of 
im  equal  to  953.1  +  0.15  (238.1)  or  988.8.  This  is  found  to  be 
64.0  at  s  =  1.617. 

18 


274 


HEAT  ENGINEERING 


,  0.202 


Moyer  states  that  Fm  may  be  found  from  the  formula 

F2  =^i[o.l72^  +  0.7o]  (8) 

Applying  this 

F2  =  0.031  [o.!72~j^  +  0.70]  =  0.195  sq.  ft. 

This  rule  would  therefore  make  the  nozzle  with  slight  under- 
expansion  and  the  effect  of  this  would  be  to  reduce  the  velocity 
by  the  percentage  shown  by  curve  of  Fig.  130  given  by  Moyer  in 
his  Steam  Turbines.  This  figure  shows  that  the  effect  of  under- 

L 


-£-6 

_S    *. 


25        20        15        10         5 

%  of  Under  Expansion 


5          10        15         20 
%  ol  Over  Expansion 


Fig.  130. — Moyer's  curves  showing  loss  due  to  variation  in  area  of  mouth. 

expansion  is  less  than  that  of  overexpansion.     In  the  problem 
above  the  amount  of  under-expansion  by  this  simple  rule  is 

202  -  195 


202 


or  3J«2  per  cent. 


The  effect  of  this  is  to  produce  practically  no  change  in  the  com- 
puted velocity. 

The  length  of  the  nozzles  is  usually  four  or  five  times  the 
diameter  of  the  throat  regardless  of  the  amount  of  flare.  If 
however  the  flare  is  over  6  deg.  it  would  be  well  to  make  the 
divergent  part  a  curve  so  as  to  support  the  steam  pressure  and 
accelerate  the  velocity.  With  a  great  flare  there  is  no  support 
for  the  steam  pressure  and  a  lower  velocity  results. 

If  the  pressure  p^  is  121.1  Ibs.  absolute,  the  above  computations 
show  that  this  is  above  the  critical  pressure  and  the  mouthpiece 


STEAM  NOZZLES,  INJECTORS,  STEAM  TURBINES       275 

is  to  be  of  the  orifice  or  converging  nozzle  form.  In  this  case 
the  smallest  section  is  at  the  end.  If  the  length  of  this  orifice  is 
made  8^2  it  might  be  well  to  assume  a  friction  factor  of  4  or 
5  per  cent,  giving: 


w2  =  223.7V(H91.1  -  1168.1)0.96  =  1050  ft.  per  sec. 

v  for  121.1  Ib.  and  *  =  (1168.1  +  0.92)  =  1169  is  3.605 

10  X  3.605 


J050 


=  0-0343  sq.  ft. 


In  figuring  the  discharge  from  a  nozzle  or  orifice  when  p2  is  less 
than  0.57pi  the  method  is  to  compute  the  discharge  through  the 
smallest  area  as  if  the  critical  velocity  existed  at  that  point,  the 
velocity  being  computed  for  the  drop  to  the  critical  pressure  and 
the  volume  taken  for  that  point.  This  method  is  used  regardless 
of  how  much  less  p%  is  than  0.57pi.  All  that  is  meant  is  that  the 
pressure  at  this  small  section  is  the  critical  pressure.  This  is 
always  true  for  orifices  while  for  nozzles  a  value  of  p2  different 
from  the  one  for  which  F2  was  designed  will  cause  a  change  in 
velocity  due  to  the  effect  of  under-  or  over-expansion,  although  the 
weight  discharged  will  remain  constant. 

INJECTORS 

The  first  application  of  the  discharge  from  nozzles  is  to  the 
injector.  This  machine  was  invented  by  Henri  Jacques  Giffard 
in  1858  although  the  same  principle  had  been  applied  by  others 
for  the  raising  of  water  by  jets.  Giffard's  injector  was  intro- 
duced into  various  countries  and  through  changes  suggested  by 
its  application  it  was  improved  by  a  number  of  inventors.  One 
of  the  modern  forms  of  injectors  is  shown  in  Fig.  131.  This  is 
the  Seller's  Improved  Self-acting  Injector.  Steam  enters  at  A 
and  to  start  the  apparatus  the  handle  B  is  drawn  back  a  slight 
distance  so  that  steam  may  enter  the  annular  space  C,  the  plug  D 
preventing  any  entrance  at  the  center.  The  combining  tube  E 
has  openings  F  and  G  leading  into  the  space  H  which  is  con- 
nected to  the  atmosphere  by  the  valve  /  which  may  be  held  down 
by  the  cam  J.  Steam  rushing  through  the  small  space  at  the 
end  of  C  produces  a  high  velocity  and,  due  to  the  fact  that  the 
space  H  is  at  atmospheric  pressure  since  the  valve  /  opens  to  the 
atmosphere,  the  pressure  at  the  end  of  C  may  be  below  that  of 
the  atmosphere  due  to  the  over-expansion.  This  together  with 


276  *HEAT  ENGINEERING 

the  entrainment  of  air  by  the  steam  jet  gradually  sucks  the  air 
from  K  producing  a  vacuum  in  this  space,  thus  drawing  water 
into  the  chamber.  This  water  meets  the  steam  issuing  from  C, 
and  entering  the  tube  E  the  water  passes  through  the  openings 
into  H  and  finally  overflows  into  the  atmosphere  as  the  cam  / 
is  raised.  When  water  appears  at  L  the  handle  B  is  drawn  back, 
and  steam  entering  the  center  imparts  such  a  velocity  to  the 
water  that  when  the  velocity  head  is  changed  into  pressure  head 
by  the  diverging  delivery  tube  M  there  is  sufficient  pressure  to 
move  the  check  valve  A"  and  allow  the  water  to  enter  the  boiler 
feed  pipe  0. 

The  function  of  the  steam  nozzle  C  is  to  lift  the  water  to  the  in- 
jector while  that  of  the  steam  nozzle  P  is  to  give  sufficient  steam 


FIG.  131. — Seller's  improved  self-acting  injector. 

to  mix  with  this  water  in  the  combining  tube  E  so  that  a  high 
velocity  will  occur  at  the  throat  Q  where  the  steam  is  completely 
mixed  with  the  water.  The  function  of  the  diverging  delivery 
tube  is  to  reduce  this  high  velocity  and  change  this  kinetic  energy 
into  pressure  or  potential  energy.  The  valve  R  opening  inward 
is  used  to  allow  extra  water  to  enter  H  and  so  mix  with  the  steam 
through  F  and  G  when  the  pressure  in  the  combining  tube  is 
less  than  atmospheric  pressure,  due  to  a  lack  of  water  at  K,  as 
the  plug  valve  S  may  not  be  opened  sufficiently.  An  excess 
pressure  in  H  caused  by  too  much  water  for  the  steam  is  shown 
by  a  slight  leakage  at  the  overflow  L.  Thus  the  tube  C  will 
restart  the  injector  and  the  valves  R  and  /  care  for  a  deficiency 


STEAM  NOZZLES,  INJECTORS,  STEAM  TURBINES       277 


or  excess  of  water  which  if  not  taken  account  of  might  cause  a 
break  in  the  proper  action  of  the  injector.  For  these  reasons  this 
is  known  as  a  restarting  self-acting  injector. 

The  important  parts  of  the  injector  are  the  steam  nozzle, 
the  combining  tube  and  the  delivery  tube.  The  forms  of  these 
have  been  the  result  of  much  experimentation.  The  steam  noz- 
zle will  be  discussed  first. 

Certain  experiments  are  described  by  Kneass  in  his  excellent 

treatise,  "  Practice  and  The- 
ory of  the  Injector."  One  of 
the  first  things  he  describes 
on  the  steam  nozzle  is  a  set 
of  photographs  of  their  differ- 


FIG.  132. — Discharge  from  nozzles 
showing  action  of  steam,  according  to 
Kneass. 


Steam  Nozzle         CombiningTube 


FIG.    133. — Exploration    curve  of 
Kneass. 


ent  forms.  In  the  first  one,  the  discharge  from  a  converging 
orifice  shows  a  sudden  enlargement  at  the  face  and  considerable 
disturbance,  although  this  is  not  so  great  as  that  in  the  dis- 
charge from  a  diaphragm  in  a  mouthpiece.  In  this  there  are  a 
great  number  of  cross  currents. 

The  disturbance  in  the  straight  taper  is  not  so  extensive  and  in 
the  discharge  from  the  curved  divergent  nozzle  it  is  practically 


278  HEAT  ENGINEERING 

eliminated.  The  enlargement  in  the  first  two  jets  is  due  to  pres- 
sure which  still  exists  in  the  steam  while  in  the  last  two  cases  this 
pressure  has  been  utilized  in  driving  the  steam  forward  so  that  by 
the  time  the  end  of  the  nozzle  is  reached  the  pressure  has  been 
utilized.  These  indicate  the  form  best  suited  for  the  purpose. 
To  study  what  takes  place  in  the  injector  he  then  explored  the 
center  of  a  nozzle  and  combining  tube  in  the  manner  described 
before  and  obtained  the  pressure  curve  shown  in  Fig.  133.  In 
this  it  is  seen  that  the  converging  part  is  rather  longer  than  has 
been  described  but  the  critical  pressure  is  seen  to  occur  at  the 
throat.  It  will  be  observed  that  the  end  of  the  orifice  is  reached 
before  the  pressure  is  reduced  to  atmospheric  pressure  but  this  con- 
tinues to  fall  to  a  lower  point  and  in  the  combining  tube/there  is  a 
strong  vacuum,  which  is  gradually  decreased,  reaching  atmospheric 
pressure  before  the  throat  of  the  delivery  tube  is  reached.  With 
no  water  present  the  steam  pressure,  marked  by  the  dotted  line, 
shows  a  drop  below  atmospheric  pressure  and  then  a  rise.  These 
two  curves  show  that  the  condensing  effect  of  the  water  is  felt 
but  this  does  not  seem  to  change  the  pressure  curve  within  the 
steam  nozzle  as  the  curves  agree  almost  exactly. 

STEAM  NOZZLE  VELOCITY 

The  form  of  steam  nozzle  usually  employed  is  one  with  a 
converging  part  followed  by  a  conical  diverging  part.  The  areas 
of  the  throat  should  be  computed  to  care  for  the  necessary  steam 
and  the  mouth  should  be  figured  for  atmospheric  pressure  but  in 
order  to  find  the  amount  of  steam  for  a  given  amount  of  water 
the  velocity  of  the  steam  jet  in  the  combining  tube  is  needed  and 
for  this  the  pressure  is  assumed  to  be  4  Ibs.  absolute  or  a  vacuum 
of  22  in.  The  formula  for  the  velocities  at  these  points  would  be 


wt  =  223.  7     t!  -  it  (9) 


wm  =  223.7     (*i  -  iatm)(l  -  0.10)  (10) 


wc  =  223.7\(*i  -  *4)(1  -  0.10)  (11) 

iatm  =  heat  content  at  atmospheric   pressure    on 

same  line  as  i\ 
it     =  heat  content  at  4  Ibs.  absolute  pressure  on 

same  line  as  i\ 

WATER  VELOCITIES 

The  water  enters  the   combining   tube  under  the  absolute 
pressure  in  the  suction  tank  minus  the  lift,  friction  head,  and 


STEAM  NOZZLES,  INJECTORS,  STEAM  TURBINES       279 

the  pressure  in  the  combining  tube.  The  friction  head  is  equal 
to  a  complex  quantity  due  to  the  sudden  bends  in  the  path,  the 
sudden  changes  in  section  and  the  friction  on  the  sides.  If  it  is 
assumed  equal  to  twice  the  velocity  head,  the  following  equation 
is  true: 

^-"(1  +  2)  =  2.30(pi  -  pe)  +34.0  -  ht 

From  this 

ww  =  A/2#K[2.3(pi  -  PC)  +  34.0  -  hi]  (12) 

pi  =  pressure  in  suction   tank   above   atmosphere  in 

pounds  per  square  inch 
PC  =  absolute   pressure   in   combining   tube   taken   as 

4  Ibs.  per  square  inch 
hi  =  lift  in  feet 

34  =  feet  head  equivalent  to  atmosphere. 
Now  the  mixture  of  water  and  condensed  steam  in  the  com- 
bining tube  must  have  sufficient  velocity  to  produce  a  pressure 
equal  to  the  boiler  pressure  when  reduced  to  a  low  velocity  of 
about  5  ft.  per  second.  If  wm  is  the  velocity  of  the  mixture  in 
the  combining  tube  and  wb,  that  on  entering  the  boiler,  the  formula 
for  this  velocity  would  be  derived  from  the  Bernoulli  equation. 


_  PblM 

~ 


2g  mc      ~   2g          mb 

Neglecting  the  losses 


wm  = 

pb  =  absolute  boiler  pressure  in  pounds  per  square  inch 
PC  =  absolute  pressure  in  combining  tube  in  pounds 

per  square  inch 

mb  =  weight  of  1  cu.  ft.  mixture  at  boiler  entrance 
mc  =  weight  of  1  cu.  ft.  mixture  in  combining  tube 
wb  =  velocity  into  boilers,  for  instance  4  ft.  per  second. 
By  experiment  Kneass  found  that  the  weight  of  1  cu.  ft.  of 
mixture  in  the  combining  tube  is  about  80  per  cent,  of  the  weight 
of  water  at  the  temperature  of  the  mixture.     This  decrease  in 
density  is  due  to  the  presence  of  air  and  steam  in  the  mixture. 

mc  =  0.80  X  mt  (14) 

mt  =  weight  of  water  per  cubic  foot  at  t°  F. 

=  60  Ib.  for  first  approximation 
mb  is  taken  as  equal  to  mt. 


280  HEAT  ENGINEERING 

THEORY  OF  THE  INJECTOR 

Now  from  the  above  equations  (11),  (12)  and  (13)  the  equation 
for  the  operation  of  the  injector  is  obtained.  The  underlying 
principle  of  the  injector  is  that  of  the  impact  of  bodies :  The  sum 
of  the  various  momenta  before  impact  is  equal  to  the  momentum 
after  impact.  In  the  impact  of  the  steam  and  water  the  action 
is  in  many  directions  among  the  various  particles  of  condensed 
steam  and  water  so  that  the  momentum  after  mixture  is  not 
equal  to  the  sum  of  the  individual  momenta.  Experiment  shows 
that  it  is  about  six-tenths  of  the  sum  or  less.  An  average  value 
of  0.5  will  be  used.  If  z  Ibs.  of  water  are  assumed  to  be  lifted  by 
1  Ib.  of  steam  the  following  momentum  equation  holds: 

0.5K  +  zww]  =  (1  +  z}wm  (15) 

This  is  the  fundamental  equation  of  action  of  the  injector. 
Steam  on  account  of  its  small  density  and  on  account  of  heat 
transfer  attains  a  high  velocity  from  the  nozzle  and  after  this  it 
is  condensed  into  a  more  dense  substance,  water,  moving  at  the 
same  velocity.  This  water  may  now  be  allowed  to  impinge  on 
other  water  moving  at  a  much  slower  velocity  and  even  after  the 
high-speed  jet  has  mixed  with  eight  or  ten  times  its  weight  of 
slow  moving  water,  the  velocity  resulting  from  the  impact  is 
sufficient  to  produce  a  pressure  great  enough  to  force  the  mix- 
ture into  the  boiler.  This  will  be  easily  understood  when  it  is 
realized  that  the  steam  would  issue  from  a  boiler  with  a  velocity 
of  3200  ft.  per  second  while  water  on  account  of  its  greater 
density  would  issue  from  the  same  boiler  pressure  with  a  velocity 
of  140  ft.  per  second. 

HEAT  EQUATION  OF  THE  INJECTOR 

There  is  now  enough  data  for  the  computation  of  z  since  wc, 
ww  and  wm  can  all  be  computed  for  given  conditions.  There  is 
one  quantity  which  has  been  assumed  and  that  is  mt  which  was 
made  equal  to  60  Ibs.  This  had  to  be  assumed  for  the  first  approxi- 
mation. After  z  is  found  the  temperature  of  the  mixture  can  be 
found  by  the  equating  of  the  energy  before  mixture  to  that  after 
mixture. 


/  , 
(q'w 


A  ,„    S 

AWw   x       "    '     A'-'      '    "~w   v   '    losses  (16) 


STEAM  NOZZLES,  INJECTORS,  STEAM  TURBINES       281 

The  energy  to  be  accounted  for  in  each  pound  of  steam  is  ii 
before  it  has  acquired  any  velocity.  It  must  be  the  same  after 
it  has  attained  its  velocity  since  no  heat  has  been  added  from  the 
outside.  The  energy  in  the  water  is  that  due  to  temperature 
plus  the  kinetic  energy  and  the  same  is  true  for  the  discharge. 
These  energies  are  figured  above  water  at  32°  F.  as  a  datum  plane. 
The  terms  representing  the  kinetic  energy  are  so  small  that  they 
may  be  neglected,  giving  the  energy  equation 

ii  +  zq'w  =  (1  +  z)q'm  (16o) 

From  (16)  q'm  is  found  and  the  temperature  corresponding 
enables  one  to  find  the  weight  of  1  cu.  ft.  of  water.  If  this 
differs  much  from  60  Ibs.,  the  value  of  wm  must  be  recomputed  and 
then,  a  new  value  of  z.  Of  course  the  value  of  0.8  for  the  density 
of  the  mixture  is  a  variable  and  is  not  absolutely  known  in  design 
so  that  a  slight  variation  from  60  need  not  require  recalculation. 

STEAM  WEIGHT 

Having  z  the  quantity  of  steam  for  a  given  weight  M  per  hour  of 
boiler  feed  can  be  found.  Here  M  represents  the  weight  of  water 
taken  from  the  suction  per  hour. 

M 


3600^ 


=  weight  of  steam  per  sec. 


STEAM  NOZZLE 


Knowing  the  weight  of  steam  per  second,  the  area  at  the  throat 
and  mouth  of  the  steam  nozzle  can  be  found  from  the  formula. 


F  =  (5) 

w 

Having  the  areas  of  the  nozzle  at  the  various  points  the  length 
is  found  by  making  it  taper  to  the  proper  area  by  a  taper  of  1  in  6. 

COMBINING  TUBE 

The  combining  tube  is  to  be  designed  in  the  next  step.  The 
form  of  this  tube  is  mainly  the  result  of  experiment.  The  water 
must  be  sustained  as  it  is  struck  by  the  steam  and  condensed 
steam.  The  form  is  a  gradually  converging  tube  which  is  made 
of  a  length  of  about  18  or  20  times  the  diameter  at  the  throat 
of  the  delivery  tube.  The  end  should  be  slightly  rounded  and 


282  HEAT  ENGINEERING 

the  annular  area  should  be  equal  to  that  required  to  admit  the 
water  with  a  velocity  due  to  the  pressure  at  this  point  which  would 
probably  be  about  one-half  the  velocity  of  the  water  at  the 
point  of  lowest  pressure. 

DELIVERY  TUBE 

The  delivery  tube  is  next  designed.  The  throat  is  designed  to 
care  for  the  mixture  of  water  and  steam  and  then  the  tube  is 
made  divergent  so  as  to  cut  down  the  velocity  and  increase  the 
pressure.  The  velocity  at  the  throat  is  computed  by  a  formula 
similar  to  that  for  wm  but  different  in  that  the  pressure  is  as- 
sumed to  be  atmospheric  at  this  point,  due  to  the  opening  into 
the  overflow. 


w'd  =       g    ~        144  +  «*>  (17) 

The  area  in  square  feet  required  at  this  point  is  given  by 


3600 


(18) 


mm  Aw'd 
A  =  relative  density  =  0.8 


This  area  Fd  of  the  combining  tube  fixes  the  number  of  the 
injector  as  most  manufacturers  use  the  diameter  of  this  point  in 
millimeters  as  the  nominal  size  or  number  of  the  injector. 

Kneass  points  out  that  the  ratio  -=r  has  a  value  between  two  and 

f  d 

three. 

The  shape  of  the  delivery  tube  is  made  in  various  forms.     It 

has  been  suggested  to  reverse  the 
form  advised  by  Nagle  for  hose 
— ^l  nozzles.      Nagle    suggested    that 

—Length  f.ne  acceleration  be  made  uniform. 

FIG.  134.— Shape  for  delivery  tube.  If  the  velocity  w'd  is  to  be  reduced 

to  Wi  in  a  given  length  the  accel- 
eration is  given  by 

o  o 


:i 


wd  —  Wb       wd 


time  length  length  (19) 

Wd  +  Wb 


STEAM  NOZZLES,  INJECTORS,  STEAM  TURBINES       283 


If  this  occurs  in  a  length  20dt,  which  is  a  length  often  used  for 
the  delivery  tube,  the  following  results: 


a  = 


(20) 


The  velocity  at  any  point  at  distance  x  from  the  throat  is 
given  by  the  equation 

w'd2  —  wx2       w'd2  — 


But 


Hence 


x    fi    -  Wb*  1  -  ^*i 
"  20dtl      ~  w'dz\    ~  dx* 


(21) 


dx  = 


Now  in  general  wb  will  be  equal  to  about  4  ft.  per  second  and 

2  "I 

w\  about  140  ft.  per  second  so  that  — j-^  =          .     This  is  negli- 
gible.    Hence 

dx  =      .    dt  (22) 

\l~m 
The  solution  of  this  is  given  by  the  following  table. 


X 

1 

2 

3 

4 

8 

12 

16 

19 

19.75 

dt 

dx 

'dt' 

1.013 

1.027 

1.040 

1.058 

1.138 

1.254 

1.496 

2.118 

3.00 

DENSITY 


The  value  of  the  density  of  a  jet  is  found  by  obtaining  the 
temperature  of  the  discharge  when  the  injector  will  just  waste  and 


284  HEAT  ENGINEERING 

the  weight  of  discharge  in  a  given  time.  If  then  the  theoretical 
amount  in  this  time  and  with  the  pressure  be  divided  into  the 
actual  amount  of  water  handled,  the  square  of  this  will  give  the 
density.  If  for  instance  an  injector  lifting  16,600  Ibs.  of  water  at 
148°  F.  will  just  discharge  this  quantity  at  160  Ibs.  gauge  pressure 
when  the  area  at  the  throat  is  0.00054  sq.  ft.  the  density  may  be 
found  thus: 

The  weight  of  1  cu.  ft.  of  water  at  148°  F.  is  61.22  Ibs.     Hence 
the  head  corresponding  to  160  Ibs.  with  a  density  of  A  is 

160  X  144       376.6 


61.22A  A 

^6 


ft. 

'•jz  •          (23) 


Weight  discharged 

=  16,600  =  -=-  X  0.00054  X  3600  X  61.22A 


IMPACT  COEFFICIENT 

The  value  of  k  could  be  worked  out  after  knowing  A  and  find- 
ing the  various  velocities  if  z  is  determined  experimentally. 
This  has  been  done  and  the  values  found  vary  from  0.25  to  0.60. 

INJECTOR  DETAILS 

The  value  of  z  is  fixed  by  the  various  velocities  and  on  account 
of  the  high  velocity  of  steam  acquired  by  discharges  at  even  low 
pressures,  exhaust  steam  could  be  used  for  feeding  boilers  or  boiler 
steam  could  be  used  to  force  water  into  a  chamber  at  a  pressure 
much  higher  than  boiler  pressure.  Thus  an  injector  could  be  used 
for  the  hydrostatic  test  of  a  boiler.  The  injector  shown  in  Fig.  131 
is  a  single  jet  injector  although  double  jets  are  sometimes  used. 
The  first  jet  lifts  the  water  and  forces  it  into  the  second  nozzle  to 
be  driven  into  the  boiler.  A  lifting  injector  is  one  which  will  lift 
its  supply  and  drive  it  while  a  non-lifting  injector  is  one  which  will 
only  force  water.  In  injector  testing  the  term  maximum  capacity 
means  the  greatest  quantity  of  water  which  could  be  sent  through 
the  injector  with  a  given  steam  pressure  and  temperature  of  feed, 
while  the  minimum  capacity  is  the  least  that  will  be  discharged 


STEAM  NOZZLES,  INJECTORS,  STEAM  TURBINES       285 

without  waste.  The  range  of  capacity  is  the  difference  of  these 
expressed  as  a  percentage  of  the  maximum  capacity.  The  over- 
flowing temperature  is  the  highest  temperature  of  operation  with- 
out overflowing  when  working  at  a  given  pressure.  The  highest 
pressure  obtainable  without  wasting  is  called  the  overflowing 
pressure. 

Injectors  can  work  as  pumps  when  operating  with  other  fluids 
than  steam,  such  as  water.  In  such  apparatus  water  under  a 
great  head  acquires  a  velocity  so  high  that  it  can  lift  several  times 
its  own  quantity  through*  a  smaller  height.  The  principle  and 
equation  of  momentum  would  be  the  same  for  this  apparatus. 
An  injector  operated  by  water  is  known  as  a  jet  pump  or  siphon. 
To  apply  the  principles  and  equations  above,  suppose  it  is 
desired  to  have  an  injector  pump  6000  Ibs.  of  water  per  hour  at 
68°  F.  into  a  boiler  at  150.1  Ibs.  gauge  pressure.  The  steam  has  a 
quality  of  0.96.  Assume  no  pressure  on  the  suction  and  a  lift 
of  6  ft. 

pt  =  0.57  X  164.8  =  93.8 

wt  =  223.7V1160.8  -  1116  =  1495  ft.  per  sec. 

wm  =  223.7V(1 160.8  -  991.1)0.9  =  2760  ft.  per  sec. 

wc  =  223.7\/(1160.8  -  917.2)0.9  =  3310  ft.  per  sec. 

vt  =  4.34  cu.  ft. 

vm  =  22.86  cu.  ft. 

(for  i  =  991.1  +  17.0  =  1008.1  at  14.7  Ibs.). 

ww  =  8.02VM[2.3  X  (  -  4)  +  34  -  6  =  20.1  ft.  per  sec. 


w     -  8  02    /(164.8  -  4)144~T~42 
Wm       8'°2V      60  X  0.8     '  +  60  " 


1/2(3310  +  220.1)  =  (1  +  2)175.5 

3310 -351.0  _ 
Z      351.0  -  20.1  ~  *'yi) 


1160.8  +  8.95  [36.1  +  ^  X  g£]  -  9.95^  +  ^  X 


1160.8  +  323  +  0.072  =  9.95g'm  +  6.17 

q'm  =  148.5 

tm  =  180°  F. 
mm  =  61.0 

The  value  of  mmA  of  60  X  0.8  will  not  have  to  be  changed 
as  this  is  as  close  as  the  value  of  A  will  warrant.     It  will  be  seen 


286  HEAT  ENGINEERING 

that  the  expressions  for  the  kinetic  energies  are  so  small  that 
they  may  be  omitted. 


6000 

Mass  of  steam  per  sec.  =  Q^AA  \/  Q  nc  =  0.187  Ibs. 

X  o.yo 


Mass  of  water  in  suction  per  sec.  =  1.66  Ibs. 
Mass  of  mixture  per  sec.  =  1.847  Ibs. 


Pl  .  -4.  =  0  000543  sq  ft 

0187X2286 


1  847 

=  0.000224  sq.  ft. 


169.6  X61.0X  0.8 
0.000543       .  . 


Fd       0.000224 
This  is  between  2  and  3. 


0.7854 
The  size  is  therefore  No.  5. 

dt  =  0.314  in. 
dm=  0.60  in. 


•°00224  ><  144  =0.202  in.  =5  mm. 


FIG.  135. — Shape  for  injector  tubes. 

The  table  for  the  delivery  tube  is  as  follows: 

x    =          0  0.202  0.404  0.606  0.808  1.616  2.424  3.232  3.838 

dx  =          0.202    0.204  0.237  0.205  0.207  0.228  0.254  0.302  0.443 

The  length  of  the  delivery  tube  is  3.8  in. 

The  length  of  the  combining  tube  is  18  X  0.202  =  3.7  in. 

The  length  of  the  entrance  to  nozzle  =  2^  X  0.314 

=  0.78  in. 
The  length  of  the  diverging  part  of  nozzle  =  4  X  0.312 

=  1.25  in. 
This  is  shown  in  Fig.  135. 


STEAM  NOZZLES,  INJECTORS,  STEAM  TURBINES       287 


STEAM  TURBINES 

The  steam  turbine  is  operated  by  the  force  exerted  when  a 
steam  jet  strikes  against  moving  blades.  If  a  jet  of  a  certain 
cross-section,  discharging  m  Ibs.  of  steam  per  second  at  a  velocity 
of  wa  strikes  a  blade  which  is  moving  in  the  direction  of  the  jet 
with  a  velocity  wb}  the  steam  is  reduced  to  this  velocity  in  the 
direction  of  the  jet.  The  acceleration  of  this  substance  is 


(24) 


if  it  is  assumed  that  this  action  takes  place  in  At  seconds.     During 

this  time  the  amount  of 

mass  acted  upon  is  mAt. 

For  although  all  of  the 

fluid     leaving     the    jet 

would  never   strike  the 

one  vane  considered,  in 

all   forms  of  apparatus 

there    are    other   vanes 

which  come  into  line  so 

that  the  amount  to  be 

considered    on  the  one 

vane   is  the  mAt  which 

strikes  this  vane  (if  desired  At  would  depend  on  the  number  of 

vanes  passing  the  jet  per  second). 

The  force  exerted  due  to  this  decrease  of  velocity  is  given  by  the 
general  formula 

P  =  ma  (25) 

Wa   —   Wb  f  ^ 

—77- —  =  m(wa  —  Wb) 


FIG.  136. — Jet  impinging  on  vanes. 


This  expression  is  in  absolute  units  of  force,  poundals  in  the 
English  system  or  dynes  in  the  French  system.  To  change  it  to 
pounds  or  grams  the  expression  must  be  divided  by  g  giving 

m(wa  -  wp) 


If  wb  is  made  zero  this  becomes 


mwc 
Q 


(26) 


(27) 


In  this  latter  case  there  is  no  work  done  as  the  blade  is  sta- 


288  HEAT  ENGINEERING 

tionary.  The  force  is  called  the  force  of  impact  or  the  impulse  of 
the  jet.  The  nozzle  of  the  j  et  must  feel  a  force  equal  to  this  as  some 
force  must  be  required  to  produce  this  flow.  This  is  called  the 
reaction  of  the  jet.  The  word  impulse  refers  to  the  force  exerted 
"by  the  jet  when  it  strikes  a  body  while  reaction  refers  to  the  force 
on  the  body  from  which  the  jet  issues  in  virtue  of  which  the  jet 
exists.  Since  in  the  case  of  the  stationary  blade  there  is  no  work 
done,  there  is  no  abstraction  of  energy  and  the  velocity  of  the 
steam  should  not  decrease.  An  investigation  made  on  water  jets 
showed  this  to  be  true.  Although  the  velocity  in  the  original 
direction  is  destroyed,  the  value  of  velocity  of  the  water  was  not 
changed;  its  velocity  in  all  directions  along  the  plane  being  equal 
to  the  original  velocity. 


FIG.  137. — Jet  impinging  on  vane  at        FIG.    138. — Actual    and    relative 
angle  with  path  of  motion.  velocities  of  the  stream  over  a  mov- 

ing blade.     Entrance  and  exit  tri- 
angles. 

In  equation  (26)  the  numerator  represents  the  change  of  ve- 
locity in  the  direction  of  motion  of  the  blade.  If  the  plane  of 
movement  of  the  moving  blade  is  not  that  of  the  nozzle  so 
that  the  actual  velocity  wa  is  inclined  at  the  angle  a  to  the  direc- 
tion of  the  motion  of  the  blade  Wb,  equation  (26)  becomes 

P  =  -(wa  cos  a  -  wb)  (28) 

If  in  addition  to  this  the  substance  is  thrown  off  from  the  blade 
with  an  actual  velocity  w'a  inclined  at  an  angle  a,  then  the  change 
in  velocity  in  the  direction  of  motion  as  shown  in  Fig.  138, 
would  be 

Wa  COS    a   —  W'a  COS   a 

and  hence  the  force  would  be 

P  —  —(wa  cos  a  —  w'a  cos  a')  (29) 


STEAM  NOZZLES,  INJECTORS,  STEAM  TURBINES      289 

If  there  were  no  motion  of  the  blade  and  no  friction,  wa  would 
equal  w'a  and  the  force  P  would  be  the  force  exerted  in  the  direc- 
tion marked  wb.  If  however  there  is  motion  of  the  blade,  wa 
doe's  not  equal  w'a  and  there  is  a  relation  between  these  and  the 
angles  of  the  nozzles  and  blades  if  the  best  conditions  prevail. 

If  wa  is  the  actual  velocity  from  the  jet  in  space  and  this  is 
directed  toward  a  blade  moving  with  a  velocity  wb,  the  motion 
of  the  fluid  relative  to  the  blade  wr  is  found  as  the  component 
of  wa  by  the  triangle  of  velocities  if  wb  is  the  other  component. 
This  means  that  to  one  standing  on  the  blade  and  moving 
with  it,  the  jet  would  appear  to  come  in  the  direction  wr  at 
the  angle  @  to  the  direction  of  motion;  hence  if  the  blade  is  to 
receive  this  stream  without  impact  or  shock  it  should  be  tangent 
to  this  direction.  If  now  the  stream  is  conducted  over  the  blade 
to  the  outlet  side  and  sent  off  relative  to  the  blade  at  an  angle 
/?'  to  the  direction  of  motion  and  with  a  relative  velocity  w'T 
the  actual  velocity  of  discharge  in  space  as  the  resultant  of  w'r 
and  wb  will  be  found  to  be  w'a  at  the  angle  a. 

Of  course  it  will  be  seen  that  as  the  blade  moves  away  from  the 
nozzle,  the  jet  will  not  impinge  on  it  but  it  must  be  remembered 
that  another  blade  or  vane  will  be  moved  up  to  take  the  place 
of  the  one  shown  in  the  figure. 

The  work  done  by  the  force  P  is  equal  to  the  force  multiplied 
by  the  distance  moved  through  in  the  direction  of  P  or 

Work  =  P  X  distance  (30) 

Distance  =  wb  X  M  (31) 

Work  =  —  (wa  cos  a  -  wfa  cos  cf)(wb  X  AO  (32) 

i? 

but  m&t  =  M 

M 
.',  Work  for  M  Ibs.  =  —  (wa  cos  a  —  w'a  cos  a')wb        (33) 

I/ 

This  is  the  work  for  M  Ib.  in  any  time.     If  M  is  the  amount  per 
second  this  will  give  the  work  per  second. 
The  work  may  be  separated  into  two  parts: 

M  M    , 

—  wa  cos  aWb  and  —  w  a  cos  a  wb 


and  may  be  said  to  be  equal  to  the  work  done  at  entrance  by 
reducing  the  velocity  to  zero  minus  the  work  done  at  exit  in 
giving  the  fluid  its  actual  discharge  velocity. 


19 


290 


HEAT  ENGINEERING 


In  other  words  the  reaction  of  the  discharge  jet  at  exit  mul- 
tiplied by  the  velocity  of  the  blade  subtracted  from  the  impulse 
of  the  jet  at  entrance  multiplied  by  the  velocity  of  the  blade  at 
entrance  is  equal  to  the  work  per  second. 

This  method  of  statement  in  two  parts  is  necessary  when  a 
turbine  is  built  with  radial  flow  in  which  Wb  is  not  the  same  at 
entrance  and  at  exit  as  shown  in  Fig.  139.  In  this  case 

work  =  —  [wbwa  cos  a  —  w'bw'a  cos  a'}  (34) 

y 

In  most  cases  steam  turbines  are  of  the  axial  flow  type  in 
which  the  steam  flows  across  from  inlet  to  outlet  in  a  direction 


Velocity  of  Whirl  _J 
Entrance 


o  ^Velocity  of  Whirl 

Exit 


FIG.  139. — Radial  flow  turbine.    Veloc- 
ities wb  and  w'b  are  different. 


FIG.  140. — Triangles  of  flow 
at  entrance  and  discharge. 


parallel  to  the  axis  and  not  in  a  radial  direction.  Hence  wb  = 
w'b.  Equation  (33)  shows  that  the  work  per  pound  of  fluid  is 

work  per  pound  =  -  (wa  cos  a  —  w'a  cos  a)wb        (35) 
y 

Wb  cos  a  and  w'b  cos  a  are  the  components  of  the  actual 
velocity  of  the  jets  in  the  direction  of  motion  and  are  called  the 
velocities  of  whirl.  Hence  the  work  per  pound  is  the  difference 

on  j- 

in  the  velocities  of  whirl  at  inlet  and  outlet  multiplied  by  — . 

y 

It  will  be  important  to  remember  that  if  the  velocity  of  whirl 
at  exit  from  the  blade  is  in  the  opposite  direction  to  that  at  en- 
trance that  this  difference  is  really  an  arithmetic  sum  since  the 


STEAM  NOZZLES,  INJECTORS,  STEAM  TURBINES       291 

sign  at  exit  is  minus.  This  is  shown  clearly  by  the  value  of 
a.  In  all  cases  the  angle  is  measured  from  Wb  when  the  arrows 
point  toward  or  away  from  the  vertex  for  wb  and  wa,  or  wb  and  wr. 
The  graphical  parts  of  Fig.  138  can  be  redrawn  in  Fig.  140. 

In  this  figure  wr  may  be  greater  or  less  than  w'r  or  equal  to  it. 
If  there  is  no  drop  in  pressure  across  the  vane  and  if  there  is  no 
friction 

Wr   —   W'r 

If  there  is  no  drop  in  pressure  and  if  there  is  friction  wr  is  greater 
than  wfr  while  if  there  is  a  drop  in  pressure  there  would  be  an 
increase  in  velocity  due  to  the  addition  of  heat  energy.  Thus 


vfr  =  ^fcft  -  «  +          [l  -  y]  (36) 

This  formula  would  give  w'r  if  there  is  a  drop  of  pressure  and 
friction.  Turbines  in  which  there  is  a  drop  in  pressure  across 
the  moving  blade  are  called  reaction  turbines  or  pressure  tur- 
bines while  those  in  which  there  are  no  changes  in  the  pressures 
as  the  steam  passes  over  the  moving  blades  are  known  as  impulse 
turbines  or  velocity  turbines.  These  names  do  not  mean  that 
impulse  takes  place  in  one  form  and  reaction  in  the  other. 
Reaction  and  impulse  are  present  in  each.  These  are  only  names 
borrowed  from  hydraulic  turbines  which  have  the  meanings  at- 
tached above. 

For  impulse  turbines  with  friction 

(37) 
(37') 

Now  y  depends  on  the  velocity  of  the  steam  over  the  blades. 
The  curves  of  Fig.  141  have  been  constructed  from  data  given  by 
Moyer,  for  stationary  and  movable  blades.  S  refers  to  the  first 
and  M  refers  to  the  second. 

In  Fig.  140  there  is  no  necessary  relation  between  a,  /?  and  $' 
although  a  is  fixed  by  the  velocities  and  the  three  angles.  The 
following  trigonometric  relations  must  hold  between  the  triangles 
of  entrance  and  exit: 

wa  sin  a  /QQ. 

tan  j8  =  -  (38) 

—  Wb 


w  r  =  wr\/l  —  ^y  =  fwr 


292 


HEAT  ENGINEERING 


wr  = 
wb 


wa  sin  a 


sin  ft 
wa  cos  a  —  wr  cos  ft 

W'r    =    Wr\/l   —  y    =  fwr 
' 


(39) 

(40) 
(41) 
(42) 

(43) 

W'r  COS  ft'    =    W'a  COS   OL     —   Wb  (40') 

Thus  if  wa,  a,  ft  and  ft'  are  given  in  a  problem  the  above 
equations  must  be  satisfied  if  there  is  to  be  no  shock  and  the 
equations  give  the  values  of  the  other  quantities.  A  graphical 
solution  is  close  enough  in  most  problems  and  the  construction 


W'a    =   VVr2  H-  Wb2  +  2WbW'r  COS  ft' 

sin  a    = 


0.40 


0.30 


0.20 


0.10 


M 


500 


1000  1500  2000 

Velocity  Relative  to  Blade 


2500 


FKJ.  141. — Values  of  y  from  data  given  by  Moyer. 

of  the  triangles  of  Fig.  140  will  give  the  results  of  the  above 
equations. 

To  save  space  the  lower  triangle  of  Fig.  140  is  turned  through 
180°  and  placed  so  that  its  apex  agrees  with  that  of  the  upper 
triangle  giving  Fig.  142.  In  many  turbines  ft  and  ft'  are  made 
supplements  of  each  other  giving  Fig.  143. 

In  this  figure  a  represents  the  velocity  of  whirl  at  entrance,  b 
represents  in  a  reversed  direction  the  velocity  of  whirl  at  exit 
and  c  represents  the  velocity  of  the  blade.  Hence  to  the  scale  of 
the  figure,  the  work  per  pound  of  steam  is 

0  —  c(a  ~^~  ^ 

Q  Q 


(44) 


STEAM  NOZZLES,  INJECTORS,  STEAM  TURBINES       293 
The  energy  in  the  steam,  as  it  strikes  the  blade,  is 

I  •  •  ..'       (45) 

since  d  represents  the  velocity  of  the  jet  leaving  the  nozzle.  The 
efficiency  of  application  of  this  jet  to  the  blade  is  spoken  of  as 
the  kinetic  efficiency  and  is  equal  to 

2c(a  +  b) 

* =     ~dT 

In  this  expression  the  scale  of  the  figure  need  not  be  known 
as  this  is  a  relation  between  the  lengths. 


FIG.  142. — Diagram  for  in- 
let and  outlet  triangles  with 
vertices  at  same  point.  Fric- 
tion on  blade. 


FIG.  143.— Triangles  with  0'  the  sup- 
plement of  0.     Friction  on  blade. 


MAXIMUM  EFFICIENCY 

If  wb  or  c  of  Fig.  143  is  increased,  the  quantity  a  would  re- 
main the  same  although  b  would  be  decreased,  so  that  it  might 
increase  the  product  if  c  were  increased.  In  any  event  there 
would  be  a  change  and  there  must  be  a  velocity  which  would 
give  the  maximum  work.  To  find  this  there  are  two  cases  to 
consider,  first  if  a  is  fixed  with  /?  =  180  —  /?',  and  second  if 
j8  and  0'  are  fixed.  There  is  no  need  of  considering  the  actual 
value  of  wa  as  this  is  only  a  matter  of  scale.  Of  course  it  must 
be  of  fixed  value  in  the  discussion,  whatever  that  value  is. 


work  =  —  [wa  cos  a  —  wfa  cos  a] 


W'a  COS  a 


'r  COS  j8'   + 


work  =  —  [wa  cos  a  —  (w'r  cos  @'  + 
cos  |S  =  -  cos  /3';  w'r  =  fwr 


(35) 

(40') 

(41) 


294 


HEAT  ENGINEERING 


Hence 

where 
Now 

Hence 


work  =  — [wa  cos  a  -f-  fwr  cos  /3  — 

t? 


/  =  vi  -  y 

Wr  COS  /?   =   Wa   COS  a  —  W 

work  =  — [(1  +  f)  {wa  cos  a  — 

u 


(40) 
(47) 


In  this  the  only  variable  is  Wb,  hence  the  work  is  a  maximum 
for  the  value  of  Wb  given  by  equating  the  first  derivative  to  zero. 


d  work  1  +  /  r 

=   0   =    -         -[Wa  COS  a   —   2wb] 


j 

dwb 


g 

=    M  Wa  COS   OL 


(48) 


FIG.  a. — No  friction.  FIG.  6. — Friction. 

FIG.  144. — Triangles  of  discharge  for  maximum  efficiency  without  friction 

and  with  friction. 

This  means  that  the  velocity  of  the  blades  should  be  one- 
half  the  velocity  of  whirl.  If  /  =  0  or  there  is  no  friction,  this 
same  fact  is  true.  The  best  result  occurs  when  the  wheel  is 
moving  at  one-half  the  velocity  of  whirl. 

If  there  is  no  friction  this  will  cause  the  absolute  velocity  of 
outflow  to  be  perpendicular  to  the  motion  of  the  wheel,  while 
if  there  is  friction  there  will  be  a  discharge  at  an  angle  to  this 
perpendicular  as  is  shown  in  Fig.  144,  a  and  b.  The  work  in  this 
case  becomes: 


work  = 


cos2  a  -  }i  waz  cos2  a] 


cos2  a 


(49) 


The  kinetic  efficiency  is: 

work 

%=  ~^7~ 

2? 


COS2  a 


(50) 


STEAM  NOZZLES,  INJECTORS,  STEAM  TURBINES       295 
If  there  is  no  friction  /  =  1  and  this  becomes  : 

vjk  =  COS2  a  (51) 

The  other  case  to  be  considered  is  that  in  which  0  and  0'  are 
fixed  and  a  may  be  varied. 

W  h 

Work  =  —  (wa  cos  a  —  w'a  cos  a')  (35) 

y 

Wh 

=  —  (wr  cos  ft  +  wb  —  w'r  cos  0'  —  wb) 

tj 

=  —wr(cos  ft  -  f  cos  00  (52) 

Now  cos  0  —  /  cos  0'  is  a  fixed  quantity  and  wbwr  are  variable 
quantities. 

The  efficiency  in  this  case  is: 

work  _  2wbWr(cos  0  —  /  cos  0Q 
^"""^f  Wa2  (53) 


>62  +  wr2  +  2wbwr  cos  0  (420 

2tlWr(c080— /C0800  /C,N 

(54) 

R 

(55) 


COS  j 

or  calling  2(cos  /3  -  /  cos  00  =   fc'and  — 


+  ^  +  2  cos 


0  = 


+  ^  +  2  cos  j8  ) 


R  =  ±  I 

.'.  wr  =  wb  (56) 

and  the  triangle  at  entrance  is  isosceles. 

0  =  2a 
Hence  the  work  becomes: 

Wh^ 

work  =  — (cos  j8  —  /  cos  00 


296  HEAT  ENGINEERING 


,.T  wa  sin  a  w( 

Now  wb 


Hence  work  = 


sin  2a         2  cos  a 

a2  (cos  )8  —  /  cos 


1  +  cos  0 
Since  4  cos2«=  2(1  +  cos  0) 

(Q  f  Qt\ 

cos  p  —  j  cos  p  ; 


for  0  =  180  -  $  and  /  =  1 

(58) 


One  other  problem  of  maximum  efficiency  should  be  con- 
sidered. Suppose  that  it  is  desired  to  operate  a  given  blade 
at  a  velocity  wbj  what  should  be  the  angle  a  and  the  velocity 
wa  to  give  the  greatest  efficiency? 

From  (52)  : 

work  =  ^-r(cos  0  -  /  cos  00  (52) 

t/ 

In  this  1%  0,  /  and  0'  are  known  and  hence  the  maximum 
work  would  occur  when  wr  is  as  large  as  it  can  be  made.  Since 
residual  energy  w'a  also  increases,  the  efficiency  may  not  be  in- 
creased. This  problem  therefore  depends  on  efficiency. 

work  _  2wbWr(cos  (3  —  f  cos  ffQ  _  ,  wr 

~'  ~~  (59) 


_  _  , 

~''  ~~ 


Now  wa2  =  wb2  +  wr2  +  2wbwr  cos  0  (420 


This  is  a  maximum  when 
-^  =  0;  or  fc(^62  +  wr2  +  2^6wr  cos  0)  =  fc(2wr2  +  2^6ifr  cos  0 

dlfr 

w;r2  =  wb2  (60) 

or  the  triangle  at  entrance  is  isosceles. 
This  fixes  wa  by  the  formula  above 

wa2  =  2wbz  (1  +  cos/3), 
wb(l 

w"  -  — 


STEAM  NOZZLES,  INJECTORS,  STEAM  TURBINES       297 

This  is  in  reality  the  same  proof  as  the  former  one  since  if 
for  a  fixed  Wb  and  blade  the  triangle  is  isosceles  giving  wa, 
for  a  fixed  wa  and  blade  the  isosceles  triangle  should  be  the  most 
efficient. 

The  three  cases  have  proven  that  where  the  angle  a  is  fixed 
the  turbine  blade  must  travel  at  one -hah*  the  velocity  of  whirl, 
whether  there  be  friction  or  not,  and  that  if  the  angles  ft  and  ft' 
are  fixed  the  best  efficiency  with  either  wa  fixed  or  wb  fixed  is 
obtained  when  the  inflow  triangle  is  isosceles.  This  holds 
whether  there  be  friction  or  not.  In  the  first  case  without  fric- 
tion, the  outflow  will  be  perpendicular  to  the  movement  of  the 
blade  while  in  all  other  cases  the  line  may  be  inclined  slightly. 
If,  however,  the  speed  of  the  blade  is  desired  and  the  angle  a  is 
fixed,  then  the  highest  efficiency  is  obtained  if  the  velocity  of 
whirl  is  made  twice  the  speed  of  the  blade.  This  fixes  wa. 


IstMoveable  1 


»5 


FIG.  145. — Velocity  compounding. 

In  the  above  figures  it  has  been  seen  that  when  the  steam  is 
used  on  one  blade  the  velocity  of  this  blade  has  to  be  equal  to 
one-half  the  velocity  of  whirl  if  a  is  fixed  and  ft  and  180  —  ft'  are 
fixed.  This  means  that  the  velocity  of  the  wheel  would  be  very 
high.  To  reduce  this,  the  velocity  of  whirl  could  be  decreased 
by  decreasing  the  velocity  from  the  nozzle.  This  is  obtained  by  a 
small  drop  in  pressure  in  the  nozzle. 

If  steam  is  to  be  used  through  a  large  difference  in  pressure 
this  would  have  to  be  utilized  in  a  series  of  nozzles  and  blades. 
This  is  called  pressure  compounding.  Another  method  is  to 
use  a  series  of  movable  and  fixed  blades  as  shown  in  Fig.  145, 
utilizing  the  high  kinetic  energy  of  discharge  from  one  blade  in 
successive  blades.  This  is  known  as  velocity  compounding. 


298  HEAT  ENGINEERING 

The  velocity  diagrams  are  combined  in  Fig.  146,  in  which 
0  =  180  -  ff 
fa  =  180  -  0'i 
In  this  the  work  per  pound  is: 

work  =  -  \a  +  b  +  ai  +  bi\ 


FIG.  146. — Combined  diagram  for  velocities  in  a  turbine  with  velocity 

compounding. 


The  efficiency  is: 


work       2c(a  +  b 


(62) 


FIG.  147. — Velocity  compounding  to  give  the  best  efficiency. 

The  value  of  wb  to  make  this  or  the  work  a  maximum  is  desired. 

It  has  been  shown  that  the  last  stage  to  give  a  maximum  re- 
sult should  have  a  velocity  of  blade  equal  to  one-half  the  velocity 
of  whirl  and  hence  the  figure  to  give  the  best  result  should  be  in  the 
form  shown  in  Fig.  147,  in  which  c  is  one-half  of  the  velocity  of  whirl 


STEAM  NOZZLES,  INJECTORS,  STEAM  TURBINES       299 


in  the  last  stage.     To  find  the  value  of  c  to  properly  fit  value  of 
a  of  the  diagram,  continue  the  lines  until  they  intersect  the  lower 

£ 

line.     The  first  intercept  is  c.     The  second  one  is  7,  since  the 

slanting  line  wr  has  been  decreased  to  w'r  =  fwr.     From  similar 
triangles 

lower  intercept        wr 
c  "  fwr 


or 


intercept  =  7 
The  second  line  is  decreased  in  the  same  manner 


Hence  the  third  intercept  is  h^j  or  H,j  h-j  • 


, 
The  fourth  intercept  is  the  same  as  the  third.     Hence 


(63) 


256 


2" 


6" 


6'" 


FIG.  148.  —  Construction    for    determination    of    c   for   multistaying  with 

friction. 

If  there  were  three  velocity  stages  this  expression  would  be 


(64) 


To  construct  this  so  as  to  find  c  graphically:  lay  off  any  dis- 
tance 1—2,  Fig.  148,  which  is  called  c;  at  right  angles  lay  off  the 
distance  1—3  equal  to  unity  to  any  scale  and  3—4  equal  to  /to 
this  scale.  Draw  an  indefinite  line  from  4  parallel  to  1—2,  pro- 


300  HEAT  ENGINEERING 

ject  2  on  this  line  at  2'  and  draw  3-2'  to  5.     1-5  is  the  distance 

f*  (* 

j>     Project  5  to  5',  draw  3—5'  to  6  and  1—6  is  equal  to  ~- 
Hence        [(1  -  2)  +  (1  -  5)  +  (1  -  6  )  +  (1  -  6)]  = 


These  are  laid  off  from  3  to  6"'and  line  3—7  is  laid  off  equal 
to  a,  the  velocity  of  whirl.  If  7  and  the  last  6'"  are  joined  and 
2"  —  8  is  drawn  parallel  to  6"'  —  7  the  distance  3  —  8  will  be 
equal  to  the  correct  distance  c  for  Fig.  146.  This  same  construc- 
tion can  be  used  for  any  number  of  stages. 

To  make  use  of  the  same  angle  of  inlet  into  the  various  steps, 
the  angle  a  of  the  second  fixed  blade  is  changed  from  the  value 


FIG.  149. — Diagram  in  which  01  =  a.     Friction  on  blades. 

to  a.  This  is  shown  by  the  diagram  in  Fig.  149  where 
the  line  has  been  swung  up  to  the  original  direction.  In  this  way 
ai  is  made  the  same  as  a  and  by  this  action  a'i  has  been  made 
larger  than  before  and  the  efficiency  has  been  increased.  The 
various  kinetic  efficiencies  for  these  arrangements  of  blades 
have  been  computed  and  result  in  the  following: 

Fig.  144a  f]k  =  81  per  cent. 

Fig.  1446  r)k  =  78  per  cent. 

Fig.  147  rjk  =  73  per  cent. 

Fig.  149  rjk  =  84  per  cent. 

These  have  been  drawn  with  a  =  cos  "  10.9. 

The  axial  thrust  is  due  to  the  difference  between  the  impulse 
at  entrance  to  the  moving  blade  and  the  reaction  at  outlet  in 
the  direction  of  the  axis.  The  force  per  pound  of  steam  per 
second  is: 


STEAM  NOZZLES,  INJECTORS,  STEAM  TURBINES       301 


p  =  -  [wa  sin  a  —  w'a  sin  a'], 
y 


(65) 


Without  friction  this  difference  is  equal  to  zero  as  is  seen  in 
Fig.  144.  If,  however,  there  is  friction  this  formula  does  not 
reduce  to  zero  but  to  a  positive  quantity. 

In  the  case  of  reaction  turbines  there  is  a  static  difference  of 
pressure  between  the  two  sides  of  the  blades  which  means  an 
axial  pressure. 

The  actual  forms  of  turbines  will  now  be  examined.  The 
simplest  turbine  is  the  DeLaval  turbine. 


Pressure 


FIG.  150.  FIG.  151. 

FIG.  150. — Section  through  DeLaval  rotor. 

FIG.  151. — Sections  of  DeLaval  turbine  with  curves  of  pressure  and 

velocity. 

Fig.  150  shows  a  section  of  the  DeLaval  turbine.  This  is  an 
impulse  turbine  in  which  velocity  is  generated  in  a  single  set  of 
nozzles  A  attached  to  a  steam  chest.  The  velocity  is  utilized 
on  one  set  of  blades.  To  increase  the  power  of  the  machine  a 
number  of  nozzles  are  used.  The  angle  of  the  nozzle  relative  to 
the  plane  of  the  blade  is  made  as  small  as  possible,  as  shown  in 
Fig.  150,  so  that  the  efficiency  which  is  proportional  to  cos2  a 
is  as  large  as  possible.  The  peripheral  speed  of  the  wheel  is  very 


302 


HEAT  ENGINEERING 


great.  wa  is  equal  to  3820  ft.  per  second  if  the  drop  in  the  nozzle 
is  from  150.1  Ibs.  gauge  to  6  Ibs.  absolute.  The  speed  of  the  wheel 
for  a  value  of  /  of  0.95  and  cos  a  =  0.9  would  be 

wb  =  J£  X  0.9  X  3820  =  1719  ft.  per  sec. 

This  would  mean  16,400  r.p.m.  for  a  radius  to  the  blades  of  1  ft. 
The  pressure  drop  for  the  axial  distances  of  nozzle  and  blade  and 
the  absolute  velocity  changes  are  shown  in  Fig.  151.  This  figure 
gives  a  section  through  the  axis  and  one  parallel  to  the  axis  through 
the  blades  and  nozzle. 

The  Curtis  turbine  is  shown  in  Fig.  152.     In  this  steam  enters 
the  nozzle  at  A  and  is  discharged  against  moving  vanes.     The 


FIG.  152. — Section  through  horizontal  Curtis  turbine. 

discharge  from  these  moving  vanes  is  guided  by  stationary  vanes 
to  another  set  of  movable  vanes,  and  after  discharge  from  these 
it  is  taken  to  another  set  of  nozzles  B  and  discharges  into  a  second 
set  of  vanes.  In  the  figure  shown  there  are  five  sets  of  nozzles, 
A,  B,  C,  Z>,  and  E,  and  to  each  of  these  there  are  two  movable 
and  one  fixed  set  of  vanes.  The  pressure  drop  takes  place  in 
five  stages,  and  there  is  no  drop  in  pressure  over  the  blades. 
The  exhaust  space  F  is  connected  to  the  condenser.  This  action 
is  better  shown  in  Fig.  153  in  which  a  section  through  the  axis 


STEAM  NOZZLES,  INJECTORS,  STEAM  TURBINES      303 


and  one  parallel  to  the  axis  of  a  two-stage  turbine  are  shown  side 
by  side  following  the  method  used  by  Moyer.  In  the  figure  the 
pressure  is  seen  to  be  constant  across  the  sets  of  vanes,  the  drop 
in  pressure  and  consequently  the  gain  in  velocity  taking  place 
in  the  nozzles.  The  turbine  is  therefore  of  the  impulse  type. 
In  some  cases  the  nozzles  which  are  arranged  in  sets  extending 


Pressure 


Absolute 
Velocity 


Axial  Length 


Section 

Perpendicular 

to  Radius 

through 

Blades 


FIG.  153. — Sections  of  Curtis  turbine  with  curves  of  pressure  and  velocity. 

over  a  portion  of  the  circumference  are  separated  into  two  sets 
on  diametrically  opposite  parts  of  the  circumference  so  as  to 
balance  the  forces  in  an  axial  direction.  The  axial  thrust  must 
be  balanced  by  some  form  of  thrust  bearing  G.  An  auxiliary 
valve  is  sometimes  used  to  admit  live  steam  into  a  lower  stage 
under  heavy  loads.  In  this  turbine  holes  are  made  through  the 


304 


HEAT  ENGINEERING 


discs  to  insure  the  same  pressure  throughout  the  moving  elements 
of  one  stage. 

For  the  Rateau  turbine  the  diagrammatic  arrangement  of  parts 
is  shown  in  Fig.  154.  In  this  turbine  there  is  only  one  set  of 
blades  to  each  set  of  nozzles  and  there  is  a  drop  of  pressure  in 
each  fixed  member.  There  is  no  change  of  pressure  across  the 


Pressure 


Absolute 
Velocity 


Axial  Length 


A 


Section 

Perpendicular 

to  Radius 

through 

Blades 


FIG.  154. — Sections  of  Rateau  turbine  with  curves  of  pressure  and  velocity. 

movable  blades  so  that  this  turbine  is  also  of  the  impulse  type. 
As  will  be  seen  later  the  area  through  which  the  steam  passes 
as  the  pressure  falls  must  increase  since  the  velocity  relative 
to  the  blades  is  about  the  same  in  each  set  but  the  specific 
volume  increases.  Hence  the  length  of  the  blades  increases. 


STEAM  NOZZLES,  INJECTORS,  STEAM  TURBINES       305 

In  all  of  these  the  steam  is  admitted  to  a  portion  of  the  circum- 
ference and  as  the  steam  pressure  falls  the  portion  of  the  cir- 
cumference covered  is  greater  so  as  to  give  greater  area  for  the 
steam.  The  position  of  the  successive  nozzles  will  be  advanced 
in  the  direction  of  flow  due  to  the  advance  of  the  steam  as  it 
passes  over  the  moving  blades.  This  is  called  lead. 

The  Parsons  turbine  is  shown  in  Fig.  155  and  diagrammatically 
in  Fig.  156.  In  this  steam  is  admitted  around  the  complete 
circumference  to  a  set  of  fixed  blades.  These  discharge  on  the 
movable  set  and  these  in  turn  to  a  fixed  set  from  which  the  action 
is  repeated.  In  each  of  these  sets  of  fixed  or  movable  blades 
there  is  a  pressure  drop  on  all  blades  and  hence  this  type  will  be 
called  reaction  type  of  turbine. 


FIG.  155. — Parsons  turbine. 

In  this  turbine  the  axial  thrust  is  balanced  by  connecting 
balancing  drums  of  proper  area  to  the  proper  parts  of  the  tur- 
bine and  connecting  the  back  of  the  last  drum  to  the  space  lead- 
ing to  the  condenser.  A  thrust  bearing  is  used  to  ensure  align- 
ment. In  the  type  of  turbine  shown  in  Fig.  155  a  double  flow 
arrangement  through  the  center  A  balances  most  of  the  thrust. 
To  make  the  turbine  more  efficient  the  Curtis  element  C  has  been 
used  for  the  first  reception  of  the  steam.  The  high-pressure 
steam  is  discharged  from  nozzles  attached  to  the  steam  chest 
B.  The  steam  from  the  movable  blades  C  then  passes  to  the 
fixed  and  movable  blades  at  D  and  finally  is  discharged  at  E. 
At  this  point  the  steam  divides  into  two  parts,  one  to  blades  / 
and  the  condenser  connection  F  and  the  other  enters  the  space 
A  at  the  center  of  the  drum  A  through  holes  and  thence  to  holes 
to  the  space  from  which  the  steam  enters  a  set  of  blades  and 
finally  issues  into  the  condenser  from  the  space  H.  This  rep- 

20 


306 


HEAT  ENGINEERING 


resents  one  of  the  more  recent  forms  of  Westinghouse  Parsons 
turbine.  The  blades  are  mounted  on  drums  A,  I  and  J  of 
different  sizes  to  allow  for  the  changes  in  volume  of  the  steam. 
The  main  steam  valve  of  the  ordinary  form  of  Parsons  turbine 
admits  steam  in  puffs  controlled  by  the  governor;  while  under 
very  heavy  load  the  governor  begins  to  operate  on  a  second 


Pressure 


Absolute 
Telocity 


Axial  Length 


^  \J 

^  ? 

1 

I 

1 

1 

I 

1 

I 

Section 
Perpendicular 
to  Kuilius 
through 
JJlades 

FIG.  156. — Sections  of  Parsons  turbine  with  curves  of  pressure  and  velocity. 

valve  admitting  extra  steam  to  the  second  stage  of  the  turbine. 
The  end  thrust,  although  reduced  by  the  two  turbines  placed  on 
the  same  shaft  in  which  the  steam  passes  to  right  and  left,  may 
exist  and  for  that  reason  a  small  thrust  bearing  is  used  to  ensure 
alignment.  The  air  leakage  around  the  shaft  into  the  steam 
space  which  is  at  the  pressure  of  the  condenser  is  prevented  by 


STEAM  NOZZLES,  INJECTORS,  STEAM  TURBINES       307 

some  form  of  steam  stuffing  box.  Fig.  156  shows  how  the  pres- 
sure drops  on  both  movable  and  fixed  blades  although  the 
absolute  velocity  decreases  over  the  movable  blade.  The 
decrease  of  absolute  velocity  would  have  been  greater  on  the 
movable  blade  were  there  no  drop  of  pressure  here. 

The  great  disadvantage  of  the  DeLaval  turbine  is  the  high 
blade  speed  and  the  resulting  complications.  For  structural 
reasons  the  maximum  output  of  this  type  is  limited  to  about 
300  kw.  The  absence  of  packing  and  the  simplicity  of  the 
machine  do  not  make  up  for  this  great  disadvantage.  Between 
the  reaction  and  the  other  impulse  turbines  there  are  advantages 
on  both  sides.  The  reaction  turbines  are  very  inefficient  on  the 
high-pressure  stages,  due  to  the  leakage  at  the  ends  of  the  short 
blades;  while  in  the  impulse  turbine  packing  around  the  shaft 
between  the  discs  separating  the  stages  is  necessary.  The  com- 
bination turbine  used  by  the  Westinghouse  Company  in  their 
use  of  a  Curtis  impulse  stage  for  the  first  utilization  of  the  steam 
and  then  the  use  of  the  reaction  blading  after  a  reduction  of 
pressure  has  been  made  to  combine  the  advantages  of  each  type. 

EFFICIENCY 

The  various  efficiencies  of  the  turbine  may  be  computed. 
The  nozzle  turns  the  heat  (i\  —  22)(1  —  y)  into  kinetic  energy 
and  ii  —  q'0  heat  units  have  been  required  per  pound.  Hence 
the  overall  nozzle  efficiency  is 

=  (ii  -  z'2)(l  -  y) 

In  some  cases  there  is  a  series  of  nozzles  in  a  turbine  and 
if  ?i,  92,  #3,  etc.,  are  the  amounts  of  heat  turned  into  work  the 
average  nozzle  efficiency  would  be 

I  i  i 

gl  +  g2  +  <?3   +     •     .     • 

i,  -  q'o 

The  kinetic  efficiency  is  best  worked  out  by  a  series  of  diagrams 
and  is  equal  to 

=  2c(a  +  fr  +  a/+6'  +  .  .  .) 

These  quantities  may  be  found  graphically  or  computed  analy- 
tically after  drawing  such  a  figure  as  Fig.  147.  The  use  of 


308  HEAT  ENGINEERING 

graphical  diagrams  for  the  simplification  of  analytical  work  in 
pointing  out  relations  should  be  well  understood. 

In  the  analysis  of  reaction  turbines  these  two  efficiencies 
cannot  be  separated.  These  are  computed  together  as  will  be 
shown  in  the  computation  for  a  Parsons  turbine. 

The  loss  due  to  radiation  from  the  turbine  may  amount 
to  1  per  cent,  of  the  heat  supplied  and  the  loss  due  to  leakage 
varies  from  1  per  cent,  in  the  case  of  impulse  wheels  of  the  De- 
Laval  form  to  5  per  cent,  in  the  case  of  reaction  turbines.  The 
efficiency  of  transmission  due  to  leakage  and  radiation  may  be 
taken  together  and  called  the  efficiency  of  weight. 

rjw  =  1  —  (loss  of  radiation  +  loss  of  leakage). 

The  next  item  which  reduces  the  delivery  from  the  shaft  of  a 
turbine  is  the  friction  of  the  blades  and  discs  which  is  known  as 
windage.  This  amounts  to  about  5  per  cent,  for  the  DeLaval 
turbines,  while  for  many  other  forms  of  discs  it  may  amount  to 
10  per  cent.  This  depends  on  the  pressure  of  the  steam  against 
the  discs  and  is  computed  for  any  given  problem  by  formulae 
given  in  text  books  on  turbine  design.  In  a'ddition  to  this  loss 
there  is  loss  due  to  friction  at  the  bearings  which  may  amount 
to  1  per  cent.  In  some  turbines  there  are  oil  pumps  and  gears 
which  consume  1  or  2  per  cent,  of  the  power  exerted  on  the  blades. 
If  the  sum  of  these  be  subtracted  from  unity  the  difference  may 
be  called  the  mechanical  efficiency,  or 

t]m  =  1  —  (windage  loss  +  bearing  loss  +  gear  or  pump  loss)  . 

The  electric  generator  will  have  an  efficiency  of  rje. 
The  overall  thermal  efficiency  is  therefore 

f]t    —    fjn    X    rjk    X    f]w    X    f]m      X    rje 

If  now  the  probable  steam  consumption  per  kilowatt  hour  out- 
put is  desired,  the  quantity  is  found  as  follows: 

2546 
One  kilowatt  hour  =     ~        =  3410  B.t.u. 


Amount  of  heat  supplied  per  Ib.  steam  =  (i  —  q'0). 
Amount  of  heat  utilized  per  Ib.  steam  =  t\t  (i  —  q'o)- 

3410 
Amount  of  steam  per  kilowatt  hour  =      ,.     —  rr  * 

f]t\i  —  q  o) 


STEAM  NOZZLES,  INJECTORS,  STEAM  TURBINES       309 
SIMILARITY  OF  ACTION  OF  TURBINE  AND  ENGINE 
The    expression   for   the   nozzle  efficiency  is  — - — r- ^ — -, —' 

1  Q   O 

This  is  the  expression  for  the  efficiency  of  the  Clausius  cycle 
with  complete  expansion  except  for  the  friction  term  (1  —  y). 
Moreover,  the  area  on  the  pv  plane  representing  the  gain  of  kinetic 
energy  is  the  area  behind  the  adiabatic.  This  is  the  area  of  an 
indicator  card  with  complete  expansion.  The  theoretic  action 
of  the  steam  is  therefore  the  same  in  the  engine  and  in  the  tur- 
bine. In  the  turbine,  however,  the  cool  portions  never  come  in 
contact  with  the  hotter  steam  and  hence  there  is  no  initial  con- 
densation. For  this  reason  there  is  no  need  of  free  expansion 
and  the  toe  of  the  expansion  line  is  utilized.  Hence  high  vacuua 
are  of  value  for  turbines.  The  gain  in  the  T.-S.  diagram  extends 
over  the  whole  width  of  the  figure  in  place  of  the  partial  width 
available  in  the  engine  with  free  expansion. 

In  the  computations  for  these  machines  the  various  types  will 
be  considered. 

COMPUTATIONS  AND  DESIGN 

DeLaval  Turbine. — It  is  required  to  get  the  leading  dimen- 
sions of  a  DeLaval  turbine  to  develop  200  kw.  with  dry  steam 
at  132.5  Ibs.  gauge  pressure  and  the  back  pressure  is  2.5  Ibs. 
gauge. 

pi  =  147.2  Ibs.  abs.  ii  =  1192.2 
pt  =  83.8  Ibs.  abs.  it  =  1146.7 
pm  =  17.2  Ibs.  abs.  im  =  1034.8 

q'm  =  188.4 

wm  =  223.7 V(ti  -  i'2)(l  -  y)  =  223.7  \/(H92.2  -  1034.8)0.88 

=  2630  ft.  per  sec. 

wt  =  223.7\Ai  -  it  =  223.7 Vl  192.2-  1146.7  =  1508  ft.  per  sec. 
vt  =  5.043  cu.  ft. 

vm  =  20.75  cu.  ft.  (for  p  =  17.19  and  i  =  1034.8  +  0.12  X  157.4) 
Assume  a  =  20° 

wb  =  y2  X  2630  X  cos  20°  =  1315  X  0.9397  =  1235  ft.  per  sec. 
2630  X  0.342 


310  HEAT  ENGINEERING 

ft  =  36°  -  2'  =  36° 
2630  X  0.342 

*•"      ~0588~~ 

/  for  1500  ft.  per  sec.  =  0.9 

1  90 
Kinetic  efficiency  =  -^-  X  (0.94) 2  =  0.840 

(1192.2-1034.8)0.88 
Thermal  efficiency  of  nozzle  =  - — 1100  0 TQ^~A =  0.1382 


Radiation  and  leakage  =  1  per  cent. 

5  per  cent,  for  windage  ] 


Friction  loss 


1  per  cent,  for  bearings  \  =  8  per  cent. 


2  per  cent,  for  gears       J 
Generator  efficiency  =  95  per  cent. 
Overall  efficiency  =  0.1382  X  0.99  X  0.840  X  0.92  X  0.95 

0.1008 
=  10.08  per  cent. 

2546 
Steam   per   kw.-hr.  =  (1192.2  _  ?S.4)0.1008  =  33'8  lbs' 

Total  steam  per  hour  =  200  X  33.8  =  6760  Ibs. 
Lbs.  per  sec.  =  1.87  Ibs. 

„      ,.      ,  ,  1.87X5.043X144 

Combined  area  of  nozzle  throats  =  -  .  gr.0  —      -  = 

1508 

0.903  sq.  in. 

n   -  t-     -i  t  1.87X20.75X144 

Combined  area  of  mouths  =  —  -  ~  =  2.124  sq.  in. 


If  10  nozzles  are  used  the  areas  of  the  throat  will  be  0.0903 
sq.  in.  and  at  the  mouth  0.2124  sq.  in.  The  diameter  will  be 
0.339  in.  and  0.520  in.  respectively.  The  nozzle  will  be  cut  away 
on  an  angle  so  that  its  length  along  the  face  of  the  blade  will  be 

'  °-520  x        -  °-520  x         =  L52  in" 


Let  the  blades  be  made  as  shown  in  Fig.  157.  The  thick 
part  in  the  center  being  introduced  to  stiffen  the  blade  and  also 
to  keep  the  area  through  the  blades  of  a  constant  width.  The 
area  required  for  the  blade  passage  per  nozzle  is 

1.87X20.75X144 


8  X  1530 


°'457 


STEAM  NOZZLES,  INJECTORS,  STEAM  TURBINES       311 


The  normal  entrance  equals  this  amount  and  the  area  along 
the  sides  of  the  blades  will  be 

0.457       0.457 
sin/3  "0.588 

If  now  the  length  of  the  nozzle  is  1.52  in.  the 
height  required  to  take  the  steam  will  be 

0.778 


1 .52  —  blade  thickness 


=  0.50  in. 


This  is  about  the  height  of  the  nozzle  and  hence 
if  the  blades  are  made  1  in.  high  there  will  be  little 
chance  for  leakage. 

In  order  to  have  space  for  10  nozzles  suppose 
the  radius  of  the  disc  be  taken  as  18  in.  Then  the 
number  of  revolutions  for  this  turbine  would  be 


N  =  -t 


1235  X  60 


27T  X  1.5 
If  this  is  desired  at  8000  r.p.m. 


=  7850 


FIG.  157. 
—  Arrange- 
ment of 
blade  thick- 
ness to  give 
uniform 
p  a  s  s  a  g  e 
width. 


1235  X  60 
radius  =  2j  x  gooo  X  12  =  17.65  in. 


FIG.  158. — Velocity  diagram  for  a  two-stage  Curtis  unit  with  varying 
values  of  /. 

Curtis  Turbine. — It  is  desired  to  find  the  leading  dimensions 
for  a  Curtis  turbine  with  four  pressure  stages  between  175  Ibs. 
gauge  pressure  and  125°  F.  superheat  and  2  in.  absolute  pressure 


312  HEAT  ENGINEERING 

at  exhaust,  each  stage  to  have  a  double  velocity  stage  or  two 
moving  blades  and  it  is  required  to  have  wa  from  nozzles  on  each 
stage  the  same  and  the  velocities  wb  the  same  on  each  disc. 
The  power  to  be  developed  is  5000  kw. 

In  such  a  case  it  is  well  to  begin  by  drawing  the  diagram  for 
velocities  and  from  this  compute  the  kinetic  efficiency,  as  this 
must  be  known  before  the  pressures  of  the  various  stages  may  be 
found.  The  value  of  /  will  not  be  the  same  for  the  various  stages, 
since  the  velocity  over  the  first  movable  blades  is  about  1500 
ft.  per  second,  over  the  fixed  blade  is  about  1200  ft.  per  second, 
and  over  the  second  movable  blade  is  about  700  ft.  per  second. 
The  values  of  /  will  be  0.9,  0.87  and  0.95.  These  varying  values 
of  /  make  the  construction  of  Fig.  148  a  little  more  complex  as  / 
is  not  the  same  on  each.  Assuming  a  =  20°  and  constructing 
the  various  values  of  /  for  Fig.  148,  Fig.  158  has  been  con- 
structed. From  this 


-  0-  (ZZ  = 


This  means  that  1  —  -r\k  or  27.8  per  cent,  of  the  kinetic  energy 
of  the  jet  of  steam  remains  in  the  steam  when  it  leaves  the  vane. 

w'  2 
This  is  partially  in  the  form  of  kinetic  energy  equal  to  ~-   and 

^Q 
part  in  additional  heat  content  from  the  friction.     This  heat  is 

therefore  to  be  added  to  the  isoentropic  heat  content  iz  at  dis- 
charge from  the  nozzle  to  find  the  heat  content  at  entrance  into 
the  next  nozzle.  Whether  the  kinetic  energy  is  considered  as 
heat  or  kinetic  energy  the  effect  is  the  same  as  shown  by  equation 
(4)  of  this  chapter. 

REHEAT  FACTOR 

The  effect  of  the  heat  is  to  increase  the  quantity  available  for 
the  lower  stages.  This  is  shown  clearly  on  the  Mollier  chart. 
Steam  flowing  from  condition  1  to  condition  2  should  have  the 
heat  content  i\  —  iz  changed  into  kinetic  energy.  On  account 
of  friction  in  the  nozzle  only  (1  —  y)(i\  —  iz)  is  changed  into 
kinetic  energy  and  in  the  Curtis  turbine  y  may  be  taken  as  0.06. 
This  means  that  0.06  (ii  —  i2)  remains  as  heat  in  addition  to  i%. 
On  account  of  the  kinetic  efficiency  of  the  blades  the  heat  (1  —  tjk) 
(1  —  y)(i\  —  2*2)  still  remains  in,  in  addition  to  the  amount  which 


STEAM  NOZZLES,  INJECTORS,  STEAM  TURBINES       313 


should  remain.     Hence  the  heat    content  at   outflow  to  next 
stage  is 

2  +  yi(ii  -  it)  +  (1  -  17*)  (1  ~  2/)0'i  -  1*2) 


lout 


=    \  - 


(64) 


This  last  statement  is  equivalent  to  saying  that  the  heat  con- 
tent remaining  is  equal  to  the  original  heat  content  minus  the 
work,  which  gives  the  point  3  for  the  proper  heat  content  at  the 
pressure  of  2,  hence  the  condition  of  outflow  from  the  first  stage 
or  inflow  to  the  second  stage  is  given  by  the  point  I'  at  which  the 
pressure  is  the  same  as  at  2,  but  i  is  the  i  of  3.  The  entropy  s 


Eutropy-a 

FIG.   159. — Mollier  chart  arranged  for  investigation  of  reheat  factor. 

has  been  increased.  If  1'  —  2'  is  made  equal  to  1  —  2  and  the 
same  friction  is  assumed,  the  velocity  of  discharge  will  be  the 
same  as  that  for  the  first  stage  and  the  reheat,  as  the  distance 
3  —  2  is  called,  will  be  the  same  since  the  same  velocity  diagram 
is  used  for  this  stage,  so  that  the  kinetic  efficiency  will  be  the 
same.  This  can  be  continued  to  l"  -  2"  and  l'"  -  2'". 
Since  the  distances  1  -  2,  l'  -  2',  l"  -  2"  and  l"'  -  2"' 
are  all  the  same  the  velocities  from  all  nozzles  will  be  equal  and 
the  diagram  of  Fig.  158  is  the  same  for  all  stages.  Hence  the 
angles  of  similar  blades  will  be  equal  as  will  the  velocities  of 


314  HEAT  ENGINEERING 

the  blades.  If  the  various  pressure  lines  are  continued  back  to 
the  entropy  line  through  1  and  2  and  the  intercepts  marked 
a,  b  and  c,  it  will  be  found  that  on  account  of  the  form  of  the 
diagram,  spreading  out  as  entropy  increases  due  to  the 
properties  of  steam,  1  —  2  is  greater  than  2  —  a;  2  —  a  is  greater 
than  a  —  b ;  and  a  —  b  is  greater  than  b  —  c.  In  other  words,  5 
times  1  —  2,  which  is  the  heat  available,  is  greater  than  1  —  d, 
the  heat  which  would  have  been  available  for  a  single  expan- 
sion. The  ratio  of  the  length  1  —  2  to  the  length  — ^—  or 

o 

heat  on  adiabatic  for  one  stage  .       ,,    .   . 

-  is  called  the  reheat  factor. 

n 

The  determination  of  this  has  been  discussed  by  Edgar 
Buckingham  of  the  U.  S.  Bureau  of  Standards,  and  is  found  in 
Vol.  vii  of  the  Bulletin  of  the  Bureau  of  Standards,  Washington, 
D.  C.  It  is  also  given  in  Reprint  No.  167.  If  this  factor  is  called 
(R.  H.)  and  is  known  in  any  case,  it  is  immediately  possible  to 
find  the  length  1  —  2  for  all  stages  since 


1  -  2  =  (—^— )  (R-  H.)  (65) 

Since  1  —  3  of  Fig.  159  is  the  heat  actually  turned  into  work 
and  1  —  2  is  the  heat  applied,  the  efficiency  of  each  stage  is 

_  1  ~  3 

778  ~  1  -  2 

This  is  the  product  of  rjn  X  rjk  or  rjs. 

In  the  triangle  321'  in  the  saturated  region 


since  the  change  in  heat  along  a  constant  pressure  line  is  equal 
to  the  change  in  quality  multiplied  by  the  heat  of  vaporization. 

r 

Now  sr  —  s3  =  1'  —  3  =  (xv  —  x2)  ^ 

2-3  (xv  -  xjr 

Hence  F^3  °r  tan  8  =  7~      ~T7  ==  T 


or  the  slope  of  the  curve  of  constant   pressure  with  the  hori- 
zontal is  proportional  to  the  temperature  or 

tan  8  =  kT  (66) 


STEAM  NOZZLES,  INJECTORS,  STEAM  TURBINES       315 

Since  T  is  constant  for  a  given  pressure  in  the  saturated  region 
8  is  constant  and  these  lines  of  constant  pressure  are  straight  lines 
and  since  i  =  0  when  s  =  0  they  should  pass  through  the 
origin  although  this  is  not  strictly  true  for  after  condensation 
sets  in  the  variation  between  i  and  s  is  logarithmic  in  form. 
However  the  divergence  between  these  lines  is  so  slight  that  the 
slope  will  not  change  much  if  they  are  assumed  to  pass  through 
zero.  Buckingham  calls  attention  to  the  fact  that  a  slight 
change  in  the  lines  will  bring  this  about.  With  this  understand- 
ing as  to  the  origin 

tan  8  =  -  (67) 

s 

If  now  7  be  the  angle  between  1  —  1'  and  3  —  1'  the  tangent 
may  be  written  as 

di 
-  (68) 


The  negative  sign  must  be  used  since  di  is  negative  when  ds 
is  positive  during  expansion. 
It  is  possible  to  write 

1  -  3 

tan  7  =  —  77 Q 

1          o 

Now      1-3  =  17.  1-2 

and       r-3  =  2T^|^(1-;')1r2 
tan  8  tan  8 

.'.  tan  7  =  :; —     -  tan  6  = 
Hence 


-  77s     S 

di  r?s       i 


or 


ds        1  —  rja  s 
di  rja      ds 


1  —  r)8  s 

If  this  is  integrated  from  1  to  the  last  point  say  lv  the  following 
results  : 

iv  rja  Si 

=  f^log,Tv 

id 

(69) 


316  HEAT  ENGINEERING 


id         v 
because  tan  5  ==  —  =  -^ 


but  Sd  =  si 

.  Si  _  id 
"^  ~  F 

It  is  also  seen  that  tan  6  =  kT,  hence 

id  =  kTdsd. 


and 

Substituting  this  above 


i*  =  ii  jr  (7°) 

id      i\  Td 


i\ 

(71) 


Now  if  the  efficiency  of  the  stages  were  all  r?s,  the  total  heat 
drop  usable  on  one  stage  would  be  ij.(ii  -  id)  and  this  would 
be  equal  to 


Since 


amount  available  with  reheat 
Now  the  reheat  factor  =  ^^  available  with  no  reheat 


(72) 


For  any  given  problem  in  which  the  limiting  pressures  and 


STEAM  NOZZLES,  INJECTORS,  STEAM  TURBINES       317 

stage  efficiency  are  known  the  temperatures  may  be  found  and 
this  formula  may  be  computed.  The  temperatures  to  be  used 
are  those  of  saturation  and  in  the  superheated  regions  these 
lines  of  constant  pressure  bend  a  slight  amount  making  a  slight 
error  in  the  application  of  the  formula  but  not  enough  to  prevent 
its  use  as  a  guide,  at  least  if  the  saturation  temperature  is  used 
for  the  upper  temperature.  The  formula  is  true  for  an  infinite 
number  of  differential  steps  while  for  a  finite  number  of  definite 
steps  the  reheat  factor  would  be  slightly  different.  The  formula 
is  therefore  of  value  to  fix  the  first  value  of  1  —  2  although  this  may 
have  to  be  adjusted  in  computing  a  problem. 
The  value  of  rjs  for  this  problem  is 

r)s  =  0.94  X  0.722  =  0.679 

T  of  saturation  for  175  Ibs.  =  370.5  +  460.7  =  831.2°. 
T  of  0.98  Ibs.  abs.  pressure  =  101.1  +  460.7  =  561.8°. 

/561.8\  °-68 
'  \S3l.2>  0.23          = 

ORS/i       561'8\    "  °'68  X  °'324  " 
°'68  I1  ~  83L2/ 

The  Mollier  chart     repared  by  Marks  and  Davis  will  be  used 
for  this  problem: 

Pi  =  189.7       superheat  =  125°  F.     si  =  1.646  '^  =  1297 
pc  =      0.98     quality       =  0.82  sc  =  1.646     ic  =  918 

7?  fT  il  ~  ic       1  04  129?  ~  918 

K.H. — -^ —     =  1.U4—    — - —       -  =  95.7  =  ^l  —  £2 

Reheat  =  (1  -  0.68)  X  98.7    =  31.6  B.t.u. 

z*2  =  1297  -  98.7  =  1198.3 
iv  =  1198.3  +  31.6  =  1229.9 
iv  =  1229.9  -  98.7  =  1131.2 
ii»  =  1131.2  +  31.6  =  1162.8 
iv  =  1162.8  -  98.7  =  1064.1 

p2     =  75  Superheat  =  32°  F. 

Pi>    =75  s  =  1.684  superheat  =  95°  F. 

pz>   =22  x  =  0.97 

Pi»  =  22  s  =  1.728  superheat  6°  F. 

p2"  =  5.3  x  =  0.93 

ii>»  =  1065.9  +  31.6  =  1095.7    pv»  =  5.3  s  =  1.781  x  =  0.965 
iv  =  1095.5  -  98.7  =    996.8   p2'"  =  1.0  s  =  1.781    x  =  0.895 


318  HEAT  ENGINEERING 

The  final  pressure  desired  was  0.98  Ibs.  and  this  result  is  as 
close  as  can  be  expected.  The  value  of  ii  —  i2  is  correct  and  wa 
may  now  be  found. 

wa  =  223.7V98.7  X  0.94  =  2146  ft.  per  sec. 

This  holds  for  each  nozzle. 

The  thermal  efficiency  of  the  nozzles  and  blades  is  therefore: 

iy.(*'i  -  £)n       0.68  X  4  X  98.7 
V       i,  -  q>0         ^1297-69.2 

The  overall  efficiency  is  found  by  assuming* 

nw  =  1  -  (0.01  +  0.02)  =  0.97 

rim  =  1  -  (0.07  +  0.01)  =  0.92 

rje    =  0.96 

rjt  =  0.219  X  0.97  X  0.92  X  0.96  =  0.188 

=  18.8  per  cent. 

The   probable  steam  per  kw.-hr.  =  0  188^1227  8  =  14'8' 

The  actual  amount  guaranteed  by  the  builders  was  15.75 
Ibs.  per  kilowatt  hour. 

Total  steam  for  5000  kw.  =  5000  X  14.8  =  74,000  Ibs.  per 
hour. 

Steam  per  second  =  20.55  Ibs. 

The  nozzles  must  be  investigated  for  the  presence  of  the  throat. 

Pti  =  0.57  X  189.7  =  108 
Ptz  =  0.57  X  75  =  42.8 
pt3  =  0.57  X  22  =  12.55 
pn  =  0.57  X  5.3  =  3.02 

There  is  a  throat  in  each  nozzle  and  i  at  these  pressures  is 
found  from  the  chart  of  Marks  and  Davis. 

t(for  108.0   Ibs.  s  =  1.646)  =  1230,  104°  superheat  =  quality 

*  (for    42.8    Ibs.  s  =  1.684)  =1180,    42°  superheat  =  quality 

i(for    12.55  Ibs.  s  =  1.728)  =  1118,      0.97  =  quality 

»(for      3.02  Ibs.  s  =  1.781)  =  1059,      0.94  =  quality 

wt  =  223.7\/1297  -  1230     =  1830  ft.  per  sec. 

tiV  =  223.7\/1229.9  -  1180  =  1580  ft.  per  sec. 

wt"  =  223.7-\/l  162.6  -  1118  =  1493  ft.  per  sec. 

wv,,  =  223.7\/1095.7  -  1059  =  1350  ft.  per  sec. 


STEAM  NOZZLES,  INJECTORS,  STEAM  TURBINES       319 

The  values  of  the  specific  volumes  are  given  below: 

vt  =      4.99  cu.  ft. 

vtf  =  10.56  cu.  ft. 
Vtf  =  30.2  cu.  ft. 
vtf"  =  110.0  cu.  ft. 

vm  =      6.24  cu.  ft.  for  p  =  75,  i  =  1297  -  0.94  X  98.7  =  1204 
tw  =    18.0    cu.  ft.  for  p  =  22,       i  =  1229.9  -  93  =  1136.9 
tw  =    64.8    cu.  ft.  for  p  =    5.3,    i  =  1162.2  -  93  =  1069.2 
vm»>  =  307       cu.  ft.  for  p  =    0.98,  i  =  1095.5  -  93  =  1002.5 

The  specific  volumes  of  the  steam  leaving  the  blades  is  equal 
to  the  specific  volume  for  the  pressures  at  these  points  and  for 
the  heat  content  of  exit  plus  the  sum  of  the  friction  loss  in  the 
nozzles  and  on  the  blades.  The  friction  loss  of  the  blades  plus 
the  residual  kinetic  energy  is  equal  to  (1-iy)  or  friction  equals 

1  —  77  —  r.e. 

Wr'2 

where  r.e.  is  equal  to  —  s- 

wa* 

ID  ^       /  ^\  ^ 
From  Fig.  158  ~T  *«  (4)    =  °-065 

W  a  \  df 

Friction  loss  on  blades  =  1  -  0.722  -  0.065  =  0.213 
Total  friction  effect  =  0.06  +  0.94  X  0.213  =  0.260 
Heat  from  friction  per  stage  =  98.7  X  0.260  =  24.7  B.t.u. 

v0  =      6.51  for  p  =  75,       i  =  1198.3  +  24.7  =  1223.0 

v*  =    18.30  for  p  =  22,       i  =  1131.2  +  24.7  =  1155.9 

tV'  =    65.5    for  p  =    5.3,    i  =  1063.9  +  24.7  =  1088.6 

?V"  =  303.4    for  p  =    0.98,  i  =    996.8  +  24.7  =  1021.5 

From  Fig.  156 

wr>  =  4^  X  2146  =  690  ft.  per  sec. 

The  areas  then  at  the  various  points  in  square  inches  are  given 
as  follows: 

20.55  X  4.99  X  144 
'  =  "  183Q  "  =  8.06  sq.  in. 

20.55  X  10.56  X  144 

=  19'7°  «•  m' 


29.55  X  30.2  X  144 
'"  =  "  1493  -    =  59.70  sq.  in. 


320  HEAT  ENGINEERING 

20.55  X  110.0  X  144 
F*"  =  ~~  =  24L° 


20.55  X  6.24  X  144 
Fm  =  "2146"  =  8'58  Sq*  m' 

20.55  X  18.0  X  144 
Fm'  =  ~  2146  "  =  24'7°  Sq*  m' 

20.55  X  64.8  X  144 
Fm"  =  2146  =  89'°  Sq'  m* 

20.55  X  307.0  X  144 
Fm'"  =  '  2146  =  422*°  sq'  m' 

The  outlet  area  from  the  last  blade  of  each  stage  is  given  by  : 

20.55  X  6.51  X  144 
F0  --  -  27.9  sq.  in. 


20.55  X  18.30  X  144 

F0>  =  79.7  sq.  in. 


20.55  X  65.5  X  144 
°"  =  "  69Q  ~  =  281-0  sq.  m. 

20.55  X  303.4  X  144 
=  ~  690  ~  =  1300.0  sq.  m. 

For  the  outlet  area  from  the  blade  of  each  stage  use  specific 
volumes  at  about  one-third  the  difference  between  the  volumes 
for  0  and  m,  using  from  Fig.  156, 

3  49 

wr'   =  T7^  X  2146  =  1515  ft.  per  sec. 
4.yo 

20.55  X  6.36  X  144 
F>0  ~~  =12.4sq.m. 


20.55  X  18.17  X  144 


20.55  X  65.3  X  144 

°"  =  ~        1515        ~  =  127-6  sq-  m- 

20.55  X  306.9  X  144 

F  o>»  =  -  -,  K1  E  -  =  600.0  sq.  in. 

lolo 

For  the  outlet  from  the  fixed  blades  the  velocities  are  each 
equal  to  1040  ft.  per  second  and  the  specific  volumes  may  be  taken 
as  the  mean  of  those  in  the  last  two  cases,  giving: 


STEAM  NOZZLES,  INJECTORS,  STEAM  TURBINES       321 

Ff  =     18.4  sq.  in. 

Ff  =    52.0  sq.  in. 

Ff»  =  186.0  sq.  in. 

Ff"  =  870.0  sq.  in. 

If  the  nozzles  of  the  first  stage  be  made  of  J^  sq.  in.  cross 
section  of  mouth  there  would  be  nine  of  these  on  each  side  of  the 
turbine.  These  will  give  an  area  of  9  sq.  in.  where  8.58  have  been 
required.  If  these  are  made  %  in.  wide  the  length  of  each  on 

the  circular  face  will  be: 

iz 

=  1.95  in. 


M  sin  20C 

These  will  take  up  about  30  in.  of  the  circumference  if  there  is 
allowance  made  for  the  partitions  between  each  two  nozzles. 
The  throats  of  these  nozzles  will  each  be: 

H  X  g1^  =  0.47  sq.  in. 
or  0.75  X  0.63  in. 

For  the  second  stage  the  mouth  may  be  made  1  sq.  in.  in 
area  requiring  13  on  each  side.  If  made  1  in.  wide  the  length  will 
be 

TTITT  =  2.92  in. 


19.7 
The  throat  will  contain  26  X  2477  =  20.8  sq.  in.     If  this  is 

1  in.  wide  the  depth  will  be 

20'8 


26  X  0.34 


=2.35  in. 


These  will  take  up  about  42  in.  on  each  side. 

If  the  areas  for  the  third  stage  are  made  3  sq.  in.  each  there 
will  be  sixteen  necessary  on  each  side  and  these  may  be  worked 
out  as  before  using  1%  in.  for  the  width. 

On  the  last  stage  the  area  of  each  might  be  made  9  sq.  in. 
requiring  25  on  each  side  and  the  width  would  be  4J^  in. 
These  would  take  up  about  160  in.  on  a  side  or  320  in.  in  the 
circumference.  If  this  were  to  use  the  complete  circumference 
the  diameter  of  the  turbine  pitch  circle  would  be  102  in.  or  8  ft. 
6  in.  If  this  is  considered  too  large  the  width  might  be  made 

6  in.,  this  would  require  92  in.  diameter.     The  blades  of  the  first 
21 


322  HEAT  ENGINEERING 

movable  set  would  be  slightly  higher  than  the  width  of  the  nozzle 
mouth.     They  will  be  determined  by  their  area  at  discharge. 
These  will  have  to  be  investigated  from 

Fo  =  length  X  height  X  sin  /3. 
Now  Fm  =  length  X  height  X  sin  a. 

Hence  since  length  is  the  same  for  nozzle  and  blades 

,    .  ,  ,         ,    Fo  sin  a 
helght°  =  h~ra^ 

124       0  34 
h°     =  X 


=         IX    04  7    X  jyTT  ==  1-19  1Q» 
=  1.75  X  ~^-  X  ~  =  2.20  in. 


The  heights  of  the  fixed  blades  will  have  to  be  made  longer  than 
the  first  movable  blades  because  although  the  specific  volume 
increases  slightly,  due  to  friction,  there  is  a  marked  decrease  in 
axial  velocity  as  seen  from  Fig.  158. 


For  the  outflow  edge  of  the  second  movable  blade  the  heights 
are  given  by: 


IX   2^7  X  Q-yg  =  1.45  in. 

oci       o  04 
1.75  X    ~~  X  ~*  =  2.41  in. 


0  ^4- 
X  •—  X  ^~  =  5.98  in. 


STEAM  NOZZLES,  INJECTORS,  STEAM  TURBINES       323 

In  this  it  is  seen  that  the  blades  will  increase  from  %-in.  nozzle 
to  0.9-in.  movable  blade,  1.02-in.  fixed  blade  and  1.07-in.  second 
movable  blade  or,  to  allow  for  spreading,  say  %  in.,  1J^  in., 
\y±  in.  and  1%  in.  These  for  the  second  stage  would  be 
1  in.,  1%  in.,  1J£  in.  and  1%  in.;  for  the  third  stage 
in.,  2%  in.,  2^  in.  and  2%  in.,*  and  for  the  last  stage 
in.,  5J«2  in.,  6  in.  and  6J^  in. 

The  angles  of  the  nozzles  would  be  20°,  the  first  movable 
blade  would  have  angles  of  24°  at  entrance,  the  fixed  blade  would 
have  an  angle  of  33°  and  the  last  movable  blade  would  be  52°. 
The  outlet  angles  are  supplements  of  the  above  as  the  blades  are 
all  symmetrical. 

The  speed  of  the  wheel,  from  Fig.  158  is 

X  2146  =  433  ft.  per  sec. 


4.96 

The  speed  with  a  diameter  of  102  in.  would  be  975  r.p.m. 
If  1000  r.p.m.  be  taken  then  D  =  100  in.  This  is  a  possibility. 
The  customary  speed  is  1500  r.p.m.  showing  that  a  smaller 
diameter  is  used.  To  use  this  diameter  of  66  in.,  the  heights  of 
the  vanes  would  have  to  be  increased  by  50  per  cent.  This 
would  make  the  last  blades  9%  in.  long. 

Rateau  Turbine. — In  this  turbine  there  are  many  pressure 
stages  and  only  one  velocity  stage  to  each  pressure  stage.  If 
this  is  the  case  and  20°  is  the  value  of  a  and  the  desired  speed  of 
blades  is  about  500  ft.  per  second,  the  velocity  across  the  vanes 
will  be  750  ft.  per  second  which  gives 

/  =  0.94 

14-0  Q4 
17*  =  ~  ^—  -cos2  a  =  0.97  X  (0.94)2  =  0.857 

Vs  =  rjn  X  rjk  =  0.94  X  0.857  =  0.805 

Suppose  the  turbine  is  to  work  through  the  same  range  of 
pressures  as  the  Curtis  turbine  above: 

561. 8\0'805 


/5(jl.8\  u> 
\831.27 


0.805 


561. 8\      0.805X0.324 


/    _5bl^X 
\         831.27 


The  heat  required  to  give  a  velocity  of  500  ft.  with  a  =  20C 
and  /  =  0.94  is  found  thus: 


324  HEAT  ENGINEERING 

2wb  1000 


•       -cos^  =  094 
The  available  heat  for  this  is 


1065 


ii  -  i*  =  0.94  \£j^)     =  21.5  B.t.u. 

The  amount  required  per  stage  is 
21.5 


1.035 


=  20.4 


XT      u       f   4  1297-918        .__ 

Number  of  stages  =  --       —  =  18-7 


Use  20  stages. 

*  _  i,  =  ™^*  X  1.035  =  19.7 

wa  =  223.7\/19.7  X  0.94  =  961  ft.  per  sec. 

961  X  0.94 

Wb  =   -     —  o  —  —  =  452  ft.  per  sec. 

a 

The  heat  returned  at  each  stage  =  19.7  X  (I  -  0.805)  =  3.84 
B.t.u. 

Heat  used  =  19.7  -  3.84  =  15.06  B.t.u. 

Actual  thermal  efficiency  =  j^—y  —  69  2  =  °'245 

Overall  efficiency  =  0.245  X  0.97  X  0.90  X  0.96  =  0.205 
The  steam  consumption  per  kilowatt  hour  would  be 

041  o 

M  -  0266x11371  '  14'75  lbs' 

Parsons  Turbine.  —  In  this  turbine  the  reheat  factor  will 
have  to  be  worked  out  from  the  blade  efficiency  of  one  set  of 
blades.  In  these  blades  there  is  a  drop  in  pressure  on  each  set 
of  blades  whether  they  are  movable  or  stationary.  If  there  is 
the  same  amount  of  heat  added  on  each  set  and  if  the  velocity 
at  entrance  is  the  same  into  each  set  there  will  be  the  same  ve- 
locity of  exit.  If  for  instance,  a  from  the  first  set  of  fixed  blades 
is  equal  to  20°  and  /3  of  the  movable  blade  is  made  equal  to 
75°,  the  value  of  w&  would  be  equal  to 

wa  sin  a  r  sin  a 

Wk  =  W,  COS  a   -     - 


Hence  for  any  desired  value  of  wb,  wa  could  be  found. 


STEAM  NOZZLES,  INJECTORS,  STEAM  TURBINES       325 


Wr    = 


wa  sin  a 


r    =  f 


sin/3 
wa  sin  a 


sin/3 


(73) 
(74) 


If  now  it  is  desired  to  use  an  angle  180  —  /8'  =  a  for  the  angle 
of  discharge  and  have  the  absolute  direction  of  the  jet  at  the 
angle  a!  =  180  —  /3  =  105°,  so  that  the  same  form  of  blades, 
although  right  and  left  handed,  may  be  used  for  each  stage, 
it  will  be  necessary  for  ^ 

w'r  actual  =  wa. 


Fixed 


Moveable 


FIG.  160. — Blades  of  a  Parsons  turbine. 
Hence  wrf  above  must  be  increased  to  wa  or 


wa  - 


-  y) 


teu"  = 


(223.7) ; 


//sina\2^ 
"  V  sin  ft  I 

^.(Eg*)*] 

~  *2  =       (1  -  2/X223.7)2 


-  y) 


(75) 


(76) 


Hence  if  a,  /8,  and  ty&  are  assumed,  wa  and  ^r  may  be  found 
and  from  wr,  f  may  be  selected,  then  i\  —  it  for  the  movable 
blade. 


326  HEAT  ENGINEERING 

For  the  fixed  blade  the  steam  enters  with  the  velocity 
,         w/  sin  a       wa  sin  a 

W  a    =   — ! -£—   =   = ^—    =   Wr 

sin  |8  sin  /3 

It  leaves  this  fixed  blade  with  the  velocity  wa,  the  same  as 
that  of  the  first  blade  after  being  affected  by  friction  and  gaining 
kinetic  energy  from  heat. 


sn 


sin 


(1  -  i/)(223.7)2 

This  does  not  give  quite  the  same  value  as  the  first  expression 
on  account  of  the  difference  in  /. 

Now  the  work  done  on  this  stage  per  pound  is 

work  =  —  [wa  cos  a  —  w'a  cos  a]  (77) 

y 

wb  ri    .    sin  a  cos  /3~|  /r_ON 

=  —  wa  cos  all  -f  -• — o —  (78) 

g  sin  |8  cos  a  J 

The  heat  used  is  *i  -  iz  +  i\  -  i \  =  Q  (79) 

.  wb               f,        tana"! 
Hence  A — wacoso:   l+i ^ 

rjs  =   -     ^     _       +  ^    _  ^n-^  (80) 

For  the  angles  mentioned  above  and  Wb  =  300  ft.  per  second: 
300  300 


Wa  = 


^ 
0.34  ~  0.849 


0  *}4 
r  =  354  X  ~**  =  124 


/  =  1.00 
354  " 


V1  "  [L°°  X 


1-0.04 

/  for  the  fixed  blade  is  the  same  as  the  value  used  above,  hence 
i\  -  i\  =  2.30 


STEAM  NOZZLES,  INJECTORS,  STEAM  TURBINES       327 

300X354  X  0.94  r      ,   0.361 
Work  per  pound  per  stage  =  ~~32~16~~         L         3^73  J 

=  3400  ft.-lbs. 
=  4.38  B.t.u. 

4.38 
Stage  efficiency  =  T^Q  =  0.95  =  rja 

The  reheat  factor  may  now  be  applied  in  the  usual  manner 
for  T?S  =  0.95.     This  is,  for  the  same  pressures  as  before, 

°-311 


''      0.95  X  0324 

Total  heat  available  =  (1297  -  918)  X  1.013  =  384  B.t.u. 
Total  number  of  stages  if  velocities  are  the  same  on  each  = 

384 

^  =  83  stages. 

This  answer  is  not  given  in  correct  form  as  it  is  customary  to 
enlarge  the  drum  as  the  pressure  falls.  If  the  first  drum  is  of 
diameter  Z>i,  the  second  is  often  made 

Z>2  =  \/2  D! 
and  D3  =  \/2  D2  =  2D1 

The  speed  will  be  proportional  to  the  diameters  and  hence  if 
the  above  is  assumed  for  the  middle  portion 


i*,-  -212 

wb3  =  300V2  =  424 

4.6 
(2*1  —  iz)  for  first  stage  is  ~-  =  2.3 

a 

(ii  -  iz)  for  last  stage  is  4.6  X  2  =  9.2 

379 

Stages  in  first  portion  =  0  ..  0  0   =  55. 

o  X  *•& 

379 

Stages  in  second  portion  =   »        .  ft  =  27. 

o  X  4.t) 

379 
Stages  in  third  po-rtion  =  9       n  0   =  14. 

o  x  y.^ 

Total  number  =  96  stages.     This  is  rather  high.     A  common 
rule  for  the  total  number  of  stages  is 

AT      u       f   4  2400000  /01, 

Number  of  stages  =  -  ^  —  (81) 


328  HEAT  ENGINEERING 

In  case  above: 

,,  2400000 

Nfirst    =  -  =  54 


_  2400000 

M  second  -        (30())2 

_  2400000 

N  third     '          (424) 2 

The  other  quantities  may  be  computed  as  shown  in  previous 
problems. 

ALLOWANCE  FOR  CHANGE  OF  RUNNING  CONDITIONS 

Where  turbines  are  actually  tested  and  one  condition  is  changed 
without  changing  other  quantities  the  following  allowances  are 
found  to  hold: 

From  0°-100°  superheat,  10°  F.  increase  in  superheat 
decreases  steam  consumption  1  per  cent. 

From  100°-200°  superheat,  12°  F.  increase  in  superheat 
decreases  steam  consumption  1  per  cent. 

From  200°-300°  superheat,  14°  F.  increase  in  superheat 
decreases  steam  consumption  1  per  cent. 

For  each  1  per  cent,  moisture  the  steam  consumption  will  be 
increased  2  per  cent. 

For  an  increase  in  vacuum  from  26  in.  to  27  in.  the  steam  con- 
sumption will  decrease  5  per  cent. 

For  an  increase  in  vacuum  from  27  in.  to  28  in.  the  steam  con- 
sumption will  decrease  6  per  cent. 

For  an  increase  in  vacuum  from  28  in.  to  28%  in.  the  steam 
consumption  will  decrease  3.87  per  cent. 

For  an  increase  in  vacuum  from  28%  in.  to  29  in.  the  steam 
consumption  will  decrease  5.75  per  cent. 

For  low-pressure  turbines  the  following  corrections  may  be 
made: 

Increase  in  vacuum  from  26  in.  to  27  in.  decreases  steam  con- 
sumption 12  per  cent. 

Increase  in  vacuum  from  27  in.  to  28  in.  decreases  steam  con- 
sumption 13.75  per  cent. 

Increase  in  vacuum  from  28  in.  to  28%  in.  decreases  steam 
consumption  8.5  per  cent. 

Increase  in  vacuum  from  28%  in.  to  29  in.  decreases  steam 
consumption  11.25  per  cent. 


STEAM  NOZZLES,  INJECTORS,  STEAM  TURBINES       329 

For  pressure,  the  increase  of  pressure  of  10  per  cent,  between 
100  and  140  Ibs.  gauge  pressure  decreases  steam  consumption 
2  per  cent. 

The  increase  of  pressure  of  10  per  cent,  between  140  and  180 
Ibs.  gauge  pressure  decreases  steam  consumption  1.95  per  cent. 

The  increase  of  pressure  of  10  per  cent,  between  180  and  200 
Ibs.  gauge  pressure  decreases  steam  consumption  1.90  per  cent. 

For  low-pressure  turbines  10  per  cent,  increase  in  pressure 
decreases  the  steam  consumption  4  per  cent. 

In  any  case  if  the  values  of  the  theoretical  thermal  efficiency 
for  two  sets  of  conditions  are  computed  the  ratio  of  steam  con- 
sumption may  be  assumed  equal  to  the  ratio  of  the  efficiencies 
and  the  equivalent  steam  consumption  thus  found. 

COMBINED  ENGINE  AND  TURBINE 

Since  the  turbine  is  of  especial  value  for  low  pressures  and 
high  pressures  lead  to  certain  troubles,  steam  engines  have  been 
used  to  handle  the  steam  first,  after  which  the  steam  is  exhausted 
into  turbines  for  use  at  low  pressures.  The  turbines  using 
exhaust  steam  are  known  as  low-pressure  turbines.  In  these 
the  exhaust  steam,  at  atmospheric  pressure  from  engines  or 
other  apparatus,  is  utilized  by  the  turbine.  This  combination  of 
engine  and  turbine  has  resulted  in  very  low  steam  consumptions. 

If  the  exhaust  is  not  sufficient  at  times  to  drive  the  turbine, 
a  set  of  blades  is  placed  in  the  turbine  on  which  high-pressure 
steam  may  be  used  in  a  high-pressure  part  in  addition  to  the 
low-pressure  steam.  This  is  known  as  a  mixed  flow  turbine. 

TOPICS 
Topic  1. — Derive  the  formula 


What  is  ?/?  What  is  the  meaning  of  i2  in  the  above  expression  and  what 
would  it  stand  for  if  the  bracket  (l—y)  were  omitted?  How  is  y  found? 

Topic  2. — Explain  how  the  area  Fx  is  found  for  different  points  of  a  nozzle 
with  a  uniform  pressure  drop.  Explain  why  a  throat  exists.  What  is  the 
critical  pressure?  Explain  the  terms:  throat,  mouth,  over  expansion, 
under  expansion.  What  is  the  effect  of  the  last  two? 

Topic  3. — Explain  how  the  pressure  in  a  tube  at  a  given  section  may  be 
found  by  experiment  and  how  it  may  be  calculated.  Explain  how  this  may 
be  used  to  compute  y.  Sketch  the  form  of  apparatus  used  by  Stodola  and 
sketch  the  form  of  his  curves. 


330  HEAT  ENGINEERING 

Topic  4. — When  are  orifices  or  converging  nozzles  used?  Sketch  Rateau's 
forms.  What  fixes  the  length  of  a  nozzle  from  entrance  to  throat  and  from 
throat  to  mouth?  Give  the  steps  taken  in  the  design  of  a  nozzle. 

Topic  6. — Sketch  and  explain  the  action  of  an  injector. 

Topic  6. — Give  the  formulae  for  the  following  velocities :  (a)  at  the  throat 
of  a  steam  nozzle  of  an  injector;  (6)  at  the  mouth  of  a  steam  nozzle  of  an  in- 
jector; (c)  at  a  point  within  the  combining  tube  for  the  steam;  (d)  at  a 
point  within  the  combining  tube  for  the  water;  (e)  at  the  throat  of  the 
delivery  tube;  (/)  after  impact  in  the  combining  tube  for  the  mixture. 

Topic  7. — Write  the  two  equations  on  which  the  action  of  the  injector  is 
based.  What  do  these  equations  determine? 

Topic  8. — Derive  the  formula  for  the  delivery  tube: 

dt 

dx  = 


Topic  9. — Explain  the  action  of  a  jet  on  a  blade  and  prove  the  formulae 

p  =mw 

g 

P    =   -(WaCOS  Ct  -   Wb). 

y 

Topic  10. — Explain  the  action  of  a  jet  on  a  curved  blade.  Sketch  the 
inflow  and  outflow  triangles  and  prove  that 

work  per  pound  =  -  (wa  cos  a  —  w'a  cos  a)wb 

What  is  meant  by  velocity  of  whirl?  What  are  impulse  and  reaction 
turbines? 

Topic  11. — Sketch  the  inflow  and  outflow  triangles  and  give  all  resulting 
trigonometric  relations. 

Topic  12. — Deduce  the  condition  for  maximum  efficiency  when  a  is  fixed 
and  ft  =  180  -  ft'. 

Topic  13. — Deduce  the  condition  for  maximum  efficiency  with  a  fixed  value 
of  ft  and  ft'. 

Topic  14. — Explain  the  meaning  of  pressure  compounding  and  velocity 
compounding.  Explain  the  construction  of  a  one-stage  and  two-stage  ve- 
locity diagram  with  friction  and  without  friction.  Explain  how  to  find  the 
work  and  kinetic  efficiency  from  this. 

Topic  15. — Explain  by  diagrams  the  peculiar  features  of  the  turbines  given 
in  text.  Sketch  the  curves  showing  the  variation  of  pressure  and  velocity 
through  the  turbine. 

Topic  16. — Give  the  expressions  for  the  various  component  efficiencies  of  a 
turbine  and  the  expression  for  the  overall  efficiency.  Explain  these  and  the 
method  of  computing  them.  Explain  the  similarity  of  thermal  action  of  an 
engine  and  steam  turbine. 

Topic  17. — Sketch  a  Mollier  chart  and  on  it  show  why  a  reheat  factor 
exists  and  derive  the  expressions 


STEAM  NOZZLES,  INJECTORS,  STEAM  TURBINES       331 

tan  5  =  kT  =  - 
s 

Si 

tan  7  =  -  <r 


-L    "~~    \  m 

R.H.  =  — 


(>-£) 


Topic  18. — Outline  the  method  of  design  of  a  turbine  to  give  a  definite 
power. 

PROBLEMS 

Problem  1. — Find  the  velocity  of  discharge  at  throat  and  mouth  of  a 
nozzle  operating  with  dry  steam  between  155  Ibs.  gauge  pressure  and  70  Ibs. 
gauge  pressure.  Find  the  area  of  the  nozzle  to  care  for  30  Ibs.  of  steam  per 
minute. 

Problem  2. — Draw  a  curve  representing  the  change  of  area  from  2  sq.  in. 
to  0.25  sq.  in.  in  0.5  in.  and  then  an  enlargement  to  2  sq.  in.  in  3  in.  Find 
the  pressure  along  this  nozzle  if  the  initial  pressure  is  135  Ibs.  absolute  and 
the  steam  is  superheated  230°  F. 

Problem  3. — Find  the  size  and  shape  of  a  Rateau  nozzle  to  deliver  2500 
Ibs.  of  steam  per  hour  from  a  pressure  of  25  Ibs.  gauge  to  a  pressure  of  15  Ibs. 
gauge.  Xi  =  0.99. 

Problem  4. — Design  an  injector  to  feed  10,000  Ibs.  of  water  per  hour  into  a 
boiler  under  a  gauge  pressure  of  220  Ibs.  The  lift  is  5  ft.  Make  sketches  of 
nozzle,  combining  tube  and  delivery  tube. 

Problem  6. — An  injector  operating  with  steam  at  175  Ibs.  gauge  pressure 
is  to  force  water  into  a  boiler  in  which  the  gauge  pressure  is  275  Ibs.  How 
much  water  is  lifted  per  pound  of  steam  ? 

Problem  6. — Prove  that  it  is  possible  to  feed  a  boiler  under  120  Ibs.  gauge 
pressure  by  means  of  steam  at  25  Ibs.  absolute  pressure. 

Problem  7. — Find  the  kinetic  efficiency  of  a  single-stage,  impulse  turbine 
with  cos  a"1  =  0.93  without  friction  and  with  friction.  Find  the  efficiency  if 
0  =  45°  and  0'  =  135°. 

wa  =  2700  ft.  per  sec. 

Problem  8. — Find  the  kinetic  efficiency  of  a  two-stage  impulse  turbine  with 
cos  oTl  =0.93.  Use  no  friction  for  first  assumption.  wa  =  2700  ft.  per 
sec.  Assume  correct  values  of  /  to  suit  velocities  and  find  the  efficiency 
with  friction. 

Problem  9. — Solve  Problem  8  assuming  that  a.  =  <xi. 

Problem  10. — Find  the  reheat  factor  for  four  stages  on  a  Curtis  turbine 
when  v)k=  0.75,  rjn  =  0.90.  The  pressure  range  is  from  175  Ibs.  gauge 
pressure,  100°  F.  superheat  to  29  in.  vacuum. 

Problem  11.  Find  the  pressure  on  the  four  stages  of  Problem  10  if  the  re- 
heat factor  is  1.03.  What  is  the  thermal  efficiency?  What  are  the  various 
efficiencies?  What  is  the  probable  steam  consumption?  Find  leading 
dimensions. 

Problem  12. — Find  the  reheat  factor  for  a  twenty-stage  Rateau  turbine 
with  the  same  values  of  efficiency  and  range  of  pressure  as  that  given  in 
Problem  10. 


332  HEAT  ENGINEERING 

Problem  13. — Determine  the  leading  dimensions  of  a  200-kw.  De- 
Laval  turbine  generator  to  operate  between  150  Ibs.  gauge  pressure  and 
2  Ibs.  gauge  pressure.  Give  the  probable  steam  consumption. 

Problem  14. — Determine  the  leading  dimensions,  number  of  stages,  reheat 
factor  and  probable  steam  consumption  of  a  Parsons  turbine  to  develop  3000 
kw.  with  pressure  of  175  Ibs.  gauge  and  50°  superheat  to  a  29-in.  vacuum. 


CHAPTER  VIII 


CONDENSERS,  COOLING  TOWERS  AND  EVAPORATORS 

The  condensers  used  in  practice  are  of  different  forms.  They 
are  of  the  surface  form  when  it  is  desired  to  use  the  condensate 
and  the  cooling  water  is  improper  for  boiler  feed.  When  the 
water  used  for  cooling  is  satisfactory  the  jet  type  may  be  used. 

Fig.  161  shows  the  usual  form  of  surface  condenser.  In  this 
steam  enters  at  A  and  passes  over  the  tubes,  which  are  filled  by 
water  which  enters  at  B  and  returns  to  C  where  it  is  discharged. 


1 


FIG.  161. — Surface  condenser. 


The  baffle  plate  D  is  used  to  spread  the  steam  over  the  surface 
of  the  tubes  and  prevent  short  circuiting.  The  condensed  steam 
is  drawn  off  at  E  and  since  the  air  entrained  in  the  boiler  feed 
would  destroy  the  vacuum  as  it  collects,  means  must  be  taken 
to  remove  this  air.  The  pump  which  takes  the  condensate  from 
this  space  of  low  pressure  to  the  atmosphere  is  made  sufficiently 
large  to  handle  this  air  as  well  as  the  water  and  is  therefore  called 
an  air  pump. 

The  air  present  with  the  steam  is  a  very  small  percentage  of 
the  mixture  as  it  leaves  the  engine  or  turbine,  but  as  the  steam 
condenses  the  non-condensable  air  becomes  a  larger  part  of  the 
remaining  mixture  of  vapor  and  air.  Air  being  heavier  than 

333 


334  HEAT  ENGINEERING 

steam,  the  mixture  containing  the  air  naturally  travels  to  the 
lower  part  of  the  condenser  where  if  it  is  not  removed  fast  enough 
it  will  so  fill  the  space  that  steam  cannot  reach  the  condensing 
surface  at  this  place  and  so  the  efficiency  of  the  condenser  is 
diminished. 

Modern  practice  has  demanded  such  low  pressures  in  the  con- 
denser that  a  separate  pump  is  necessary  to  remove  the  air  alone 
from  the  bottom  of  the  condenser  at  F  while  a  small  pump  is  used 
for  the  condensed  steam  at  E.  The  air  pump  is  then  called  a 
dry  air  pump,  and  by  taking  this  air  through  a  space  cooled  by 
tubes  containing  the  coolest  water  available  the  volume  of  this 
air  is  decreased  by  the  increase  of  air  pressure  due  to  the  decrease 
of  the  vapor  pressure. 

PRESSURES  IN  A  CONDENSER 

If  pc  is  the  actual  pressure  in  the  condenser  shown  by  the 
vacuum  gauge,  this  pressure  is  equal  to  the  sum  of  the  pressure 
due  to  the  steam  and  that  due  to  the  air.  Since  the  steam  is  in 
the  presence  of  water,  the  steam  pressure  ps  is  the  saturation 
pressure  corresponding  to  the  temperature  and  the  air  pressure 
pa  is  the  difference  of  these. 

Pa    =    PC    -    Ps  (1) 

The  pressures  pa  and  ps  are  known  as  partial  pressures. 

If  now  by  passing  this  mixture  of  vapor  and  air  through  a  cold 
space  p8  is  made  smaller,  the  pressure  pa  will  be  made  so  much 
greater  that  the  volume  of  the  air  will  be  materially  decreased. 
This  requires  a  smaller  air  pump. 

The  pressure  present  in  a  condenser  with  no  air  corresponds 
to  the  saturation  temperature,  but  as  air  is  added  this  pressure 
rises,  or  for  a  given  pressure  the  temperature  of  the  mixture  will 
be  lowered  as  more  and  more  air  is  permitted  to  enter. 

The  amount  of  air  present  has  been  the  subject  of  much  in- 
vestigation. G.  A.  Orrok  in  the  Transactions  of  the  American 
Society  of  Mechanical  Engineers,  Vol.  xxxiv,  p.  713,  etc.,  has 
shown  that  although  Croton  water  contains  4.3  per  cent,  of  air  by 
volume,  this  is  decreased  to  about  0.93  per  cent,  when  heated  in 
open  heaters  to  187°  F.;  0.0151  per  cent,  of  the  0.93  per  cent,  is 
air  mechanically  mixed  and  0.916  per  cent,  is  in  solution.  This 
air  is  referred  to  its  volume  at  atmospheric  pressure.  Although 
the  feed  water  contains  0.93  per  cent,  of  air  the  water  in  the 


CONDENSERS,  COOLING  TOWERS  AND  EVAPORATORS     335 


hot  well  contains  0.269  per  cent,  of  air,  showing  that  for  eaoh 
cubic  foot  of  feed  water  0.00661  cu.  ft.  of  air  at  atmospheric 
pressure  had  been  liberated.  At  the  low  partial  air  pressure 
present  in  the  modern  condenser  this  is  increased  a  hundredfold 
in  volume.  Mr.  G.  J.  Foran  in  the  discussion  of  this  paper  gives  a 
curve  from  Castell-Evans  Physico-Chemical  tables  showing  the 
percentage  absorption  at  different  temperatures  when  the  total 
pressure  outside  is  760  mm.  This  is  given  in  Fig.  162.  Although 
this  does  not  agree  with  the  values  found  by  Orrok  it  shows  how 
the  solubility  changes  with  the  temperature. 


Absorbtion  Coefficient,  Cu.Ft.of  Dry  Air  per  Cu.Ft.oi  Water 
PPPPgggggP? 

/ 

/ 

f 

x^ 

x 

X 

,x 

x^ 

.^ 

^ 

^^ 

^ 

X 

/ 

' 

X 

x 

/ 

/ 

£        212DF 


192 


172 


152  132  112 

Temperature  in  Degrees   F 


<J2 


72 


52 


FIG.  162. — Amount  of  air  at  atmospheric  pressure  absorbed  by  1  cu.  ft.  of 
water  at  various  temperatures  given  by  Foran  from  Winkler's  data. 

Orrok  found  that  with  turbine  units  of  5000  to  20,000  kw. 
the  amount  of  air  at  atmospheric  pressure  and  temperature 
varied  from  1  cu.  ft.  per  minute  when  the  units  were  tight  to  15 
or  20  cu.  ft.  with  ordinary  leakage,  and  40  to  50  cu.  ft.  per  minute 
when  the  units  were  not  tight.  This  shows  that  such  air  is  due 
to  leakage  and  not  to  the  air  contained  in  the  feed  water.  The 
minute  holes  in  the  shells,  heads  and  expansion  joints  are  difficult 
to  detect,  but  the  saving  from  the  elimination  of  these  is  much 
greater  than  the  cost  of  that  elimination.  Although  this  leakage 
may  amount  to  a  considerable  quantity,  the  shutting  down  of  the 
air  pump  does  not  mean  an  immediate  loss  of  vacuum.  In  a 
test  mentioned  by  Mr.  Foran  a  condenser  caring  for  9000  kw.  lost 
only  0.3  in.  in  30  minutes. 


336 


HEAT  ENGINEERING 


steam  iniet  Orrok  f ound  that  there  was  too 

much  drop  of  pressure  in  the  con- 
denser proper.  To  reduce  this 
the  condensers  should  be  made 
broad  and  shallow.  The  drop 
amounted  to  about  0.2  in.  of 
mercury,  increasing  with  the  load 
to  0.6  in.  He  found  that  the 
power  required  for  the  air  pump 
condensers  amounted  to  about  25 
i.h.p.  on  the  steam  end  for  loads 
between  6000  and  10,000  kw. 
The  mechanical  efficiency  of  the 
dry  air  pumps  was  about  50  per 
cent,  and  the  volumetric  efficiency 
working  between  J^  Ib.  and  15 

Ibs.  was  very  low  indeed,  due  partially  to  the  low  value  of  the 
clearance  factor  and  to  the  large  ratio  of  compression. 


Steam 


FIG.  163. — Shell  and  partitions  of 
a  Wheeler  dry  tube  condenser. 


Condensate 

Uniflux  Contraflo 

FIG.  164. — Sections  of  uniflux  and  contraflo  condensers  in  which  steam  pas- 
sages grow  small  as  steam  condenses. 

During  these  tests  the  absolute  pressure  was  about  1%  m-  °f 
mercury  and  was  obtained  with  15°  F.  to  20°  F.  rise  in  the  cooling 


CONDENSERS,  COOLING  TOWERS  AND  EVAPORATORS    337 


To  Dry  Air  Pump 


water  at  35°  F.  The  condensed  steam  was  10°  above  the  tem- 
perature of  the  outlet  water  although  the  steam  space  might  be 
30°  above  this. 

To  better  the  cooling  effect  of  the  surface  of  condensers  the 
velocities  of  the  water  and  steam  across  the  surface  should  be 
made  high.  This  was  explained  in  Chapter  III. 

The  attempt  has  been  made  to  keep  the  tubes  partially  dry 
from  the  drip  of  the  condensate  in  the  Wheeler  Dry  Tube  Con- 
densers, Fig.  163,  by  putting  partitions  across  the  condenser. 
The  condenser  tubes  have  been  omitted  from  the  figure  so  that 
the  partitions  may  be  shown 
clearer.  This  of  course  in- 
creases the  velocity  of  the 
steam  and  so  improves  the 
heat  transmission.  This  is  a 
development  of  the  English 
contraflo  condenser.  In 
Weir's  Uniflux  Condenser,  Fig. 
164,  the  shell  is  so  made  that 
the  velocity  of  steam  remains 
nearly  constant  as  it  is  con- 
densed and  crosses  the  tubes. 
As  the  volume  decreases,  due  to 
the  condensation  of  part  of  the 
steam,  the  space  through 
which  the  vapor  passes  gradu- 
ally diminishes.  The  uniflux 
condenser  shown  at  the  left  of 
Fig.  164  is  almost  an  equivalent 
of  the  contraflow  in  a  circular 
shell,  the  partitions  of  the 
contraflow  guiding  the  steam 


Barometric  Tube 
34  ft.  Long 


FIG.  165. — Barometric  jet  condenser. 


into  a  channel  of  decreasing  area. 

When  the  condensing  water  is  suitable  for  boiler  feed  or  where 
it  is  not  necessary  to  save  the  condensate,  jet  condensers  are  often 
used  because  they  are  cheap  and  may  be  effective.  In  them  the 
water  and  steam  are  brought  into  intimate  contact.  Fig.  165 
shows  the  type  of  jet  condenser  known  as  the  barometric  counter- 
current  condenser 

In  this  steam  enters  at  A  and  meets  the  numerous  streams  of 
condensing  water  by  which  it  is  condensed,  and  falls  to  the  tail 

22 


338  HEAT  ENGINEERING 

pipe  B  and  finally  discharges  into  the  hot  well.  The  cold  water 
enters  at  C  and  discharges  through  numerous  orifices  into  the 
highest  trough  D  from  which  it  discharges  into  the  other  troughs 
in  succession.  Air  is  removed  by  a  dry  air  pump  connected  at 
E  and  all  air  has  to  pass  through  streams  of  the  coolest  water 
before  leaving  the  condenser  head.  In  this  way  p8  is  made  as 
small  as  possible,  and  pa  as  great  as  possible. 

CONDENSER  DESIGN 

The  first  point  to  be  settled  in  the  design  of  a  condenser  is  the 
temperature  of  the  cooling  water  and  the  vacuum  desired.  In 
most  cases  there  will  be  a  rise  of  about  20°  in  the  cooling  water, 
its  outlet  temperature  will  be  10°  to  30°  less  than  the  tem- 
perature in  the  condensing  space  and  the  temperature  of  the  hot 
well  will  be  less  than  the  temperature  in  the  condenser  by  about 
10°  or  20°.  A  standard  assumed  by  condenser  designers  for 
high  vacuum  work  is  a  15°  rise  in  the  cooling  water  and  a  tem- 
perature of  outlet  of  15°  below  the  temperature  corresponding 
to  the  vacuum  carried.  The  greater  the  difference  in  temperature 
between  the  water  and  the  steam,  the  more  effective  the  surface 
will  be.  If  then  ti}  the  inlet  temperature,  is  known,  t03  the  tem- 
perature at  outlet,  is  assumed  and  finally  ts  of  the  steam  is  as- 
sumed. From  this  ps  is  known,  and  then  if  the  amount  of  air 
present  is  known  pa  may  be  computed  and  from  it  pc,  the  con- 

T) 

denser  pressure  expected  on  the  condenser.     The  ratios  of   - 

PC 

will  be  about  0.8  or  0.9.  Knowing  this,  the  amount  of  water 
per  pound  of  steam  is  given  by 

(2) 


q'o  -  q'i 

where      G    =  pounds  of  cooling  water  per  pound  steam 
i0    =  heat  content  of  steam  entering  condenser 
q'h  =  heat  of  liquid  at  temperature  of  hot  well 
q'o  =  heat  of  liquid  at  temperature  of  outlet 
q'i  =  heat  of  liquid  at  temperature  of  inlet. 

If  this  is  a  jet  condenser 

q\  =  q'o  (3) 

These  may  be  taken  within  5°  of  the  temperature  in  the  head. 


CONDENSERS,  COOLING  TOWERS  AND  EVAPORATORS    339 

After  G  is  found  the  next  calculation  is  that  of  the  surface 
required  for  a  given  amount  of  steam.  In  practice  \Y±  to  2 
sq.  ft.  of  surface  are  used  per  kilowatt. 

If  W  is  the  weight  of  steam  per  kilowatt-hour  output  the  total 
weight  of  water  used  will  be 

total  water  per  hr.  =  G  X  W  X  kw.  (4) 

The  quantity  of  heat  is 

Q  =  G  X  W  X  kw.  X  (q'0  -  q'i)  (5) 

Using  the  formulae  of  Chapter  III  for  the  surface  the  following 
results : 

F  =  KmesiuAT   °r    heat  per  sq.  ft. 

In  computing  the  heat  per  square  foot  the  velocity  of  the  water 
may  be  taken  as  4  ft.  per  second. 

The  tubes  are  arranged  in  nests  so  that  the  velocity  of  the  water 
is  that  desired  and  then  the  spaces  between  the  tubes  are  made 
to  give  a  uniform  velocity  by  means  of  the  partitions. 

After  this  the  power  and  size  of  the  water  pump  may  be  com- 
puted by  assuming  a  pressure  of  say  5  Ibs.  for  the  resistance  in 
the  condenser  tubes.  The  power  and  size  of  the  air  pumps  are 
then  found. 

These  various  points  will  be  brought  out  in  the  problems  below. 

PROBLEMS     . 

Problem  1. — Suppose  that  20  cu.  ft.  of  air  per  minute  are  found  to  be 
present  in  the  exhaust  of  a  10,000-kw.  turbine.  At  a  temperature  of  75° 
in  the  condenser  chamber  and  a  pressure  of  1  in.  what  would  be  the  values  of 
ps  and  pa]  and  what  would  be  the  amount  of  air  to  be  cared  for  by  an  air 
pump  if  the  air  is  cooled  to  60°  F.  before  entering  the  cylinder?  Steam  per 
kilowatt  hour  =  14  Ibs. 

Steam  pressure  at  75°  =  ps  =  0.4289  Ib.  per  sq.  in. 

Total  pressure  =  pc  =  0.49  Ib.  per  sq.  in. 

Partial  air  pressure  =  pc  —  PS  =  0.0611  Ib.  per  sq.  in. 

Steam  pressure  at  60°  =  0.2561  Ib.  per  sq.  in. 

Air  pressure  =  0.2339  Ib.  per  sq.  in. 

It  will  be  seen  that  by  the  cooling  of  this  air  the  pressure  has  been  increased 
fourfold. 

Volume  of  air  per  min.  =  20  X  9^339  ^  1283  cu-  ft- 


340  HEAT  ENGINEERING 

The  pressure  of  the  air  in  the  dry  air  pump  is  raised  from  54  Ib.  to  15  Ibs. 
and  the  clearance  factor  and  leakage  factor  would  be  computed  as  in  the 
case  of  the  air  compressors  of  Chapter  IV. 


Displacement  =  ^learance  factor  X  leakage  factor  (7) 

n  PlV,          V  ~  W~n~) 

n  -  1  leakage  factor  33000 

Problem  2.  —  Find  the  amount  of  cooling  water  for  the  condenser  of 
Problem  I  if  the  inlet  water  is  at  50°  F. 

U  =  50° 
tc  =  75° 
t0  assumed  65° 
th  =  55° 

Value  of  i15  =  43.1  +  1049.2  X  0.895  =  983.1 
(0.895  was  assumed  from  turbine  problem  in  Chap.  VII) 
983.1  -  23.1 


33.1-  18.1 

Total  weight  water  =  10,000  X  14  X  64  =  8,950,000  Ibs.  per  hr. 
Total  volume  per  minute  =  2400  cu.  ft.  or  17,800  gal.  per  min. 

Problem  3.  —  Find  the  surface  required  for  the  condenser  above  using 
admiralty  tubing  and  assuming  clean  tubes  with  a  water  velocity  of  4  ft. 
per  second. 

A*i  =  75  -  50  =  25 


75  -  65  =  10 


[ 


(25)*  - 
630  X  1.0  X     ~     0.98Vi 


Total  heat  per  hr.  =  8,950,000  X  15  =  670  X  F  X  16.4 

F  =  12,200  sq.  ft. 
Increasing  this  20  per  cent,  gives 

F  =  14,600  sq.  ft.  or  1.46  sq.  ft.  per  kw. 

In  the  N.  Y.  Edison  Plants  1.1  sq.  ft.  are  used  per  kilowatt  in  one  of 
their  new  turbines.  At  the  Interborough  Station  in  New  York  1.67  sq.  ft. 
are  used,  and  at  the  Fisk  Street  Station  in  Chicago  2.08  sq.  ft.  are  used. 

The  cost  of  operating  the  auxiliaries  of  condensers  amounts  to 
3J£  per  cent,  of  the  power  of  the  engine  or  turbine,  about  K  Per 
cent,  being  taken  for  the  air  pump  and  3  per  cent,  for  the  cir- 
culating pump. 


CONDENSERS,  COOLING  TOWERS  AND  EVAPORATORS    341 

In  the  case  of  turbines  a  low  vacuum  can  be  used  because  the 
toe  of  the  diagram  is  used,  the  expansion  of  the  steam  being  of 
value  to  the  lowest  point.  On  account  of  the  steam  engine  be- 
ing unable  to  use  the  toe  of  the  expansion  diagram,  due  to  the 
enormous  volume  required,  the  gain  in  power  for  part  of  the 
volume  by  decreasing  the  pressure  is  not  enough  to  pay  for  the 
greater  expense  in  operating  the  pump  and  the  reduction  in  the 
temperature.  This  was  explained  in  Chapter  II. 


COOLING  TOWERS 

Where  cooling  water  is  not  obtainable  as  in  the  center  of  a 
large  city  and  it  is  desired  to  operate  the  apparatus  condensing, 


FIG.  166. — Cooling  tower. 

cooling  towers  are  used.  These  are  of  different  forms.  In  all 
of  them  air  is  allowed  to  come  in  contact  with  the  warm  condens- 
ing water  which  has  been  divided  by  some  device  into  fine  streams 
or  sheets.  In  some  cases  the  water  is  allowed  to  trickle  over 
mats  or  screens  of  wire  or  is  discharged  over  vitrified  tiles  by  a 
distributing  head.  In  other  cases  the  water  is  discharged  over 
boards.  The  general  plan  is  to  get  the  water  into  thin  sheets  or 
drops  so  as  to  expose  a  large  area  to  the  action  of  the  air.  To 
bring  the  air  in  contact  with  the  water  it  is  forced  through  by  a 
fan  or  by  a  flue  or  chimney  built  on  top  of  the  tower  to  produce 
a  draft  by  the  warm  air  which  sucks  fresh  air  into  the  bottom  of 


342  HEAT  ENGINEERING 

the  tower.  This  air  becomes  heated  from  the  warm  water  and 
has  its  moisture  content  so  increased  that  the  evaporation  of  the 
water  to  satisfy  this,  added  to  the  heat  required  to  warm  the  air, 
so  cools  the  water  that  it  reaches  the  bottom  of  the  tower  at  a 
temperature  suitable  for  the  condenser. 

Fig.  166  illustrates  one  form  of  tower.  The  fans  A  are  driven 
by  an  engine  or  motor  and  force  the  air  through  the  wire 
screens  B,  called  mats.  The  water  enters  the  troughs  C-C  from 
the  pipe  D  and  is  distributed  into  tubes  E  which  have  their  upper 
part  removed,  forming  troughs  which  overflow  and  distribute 
their  water  over  the  mats.  By  continuing  the  casing  30  or  40  ft. 
above  the  troughs  there  would  be  sufficient  chimney  effect  to 
draw  considerable  air,  although  for  maximum  capacity  fans  are 
required.  Cooling  towers  take  about  5  per  cent,  of  the  power  of 
the  apparatus  being  operated  by  the  condenser,  2  per  cent,  for 
the  fan  and  3  per  cent,  for  circulating  the  water. 

The  amount  of  moisture  that  will  be  taken  up  by  air  depends 
on  the  amount  of  moisture  present  in  the  air.  All  liquids  dis- 
charge particles  into  any  space  around  them  forming  a  vapor  of 
the  substance  in  the  space  above  the  liquid.  Particles  also  fall 
back  from  the  space  into  the  liquid.  The  number  of  particles 
leaving  the  liquid  is  greater  than  the  number  falling  back  until 
the  pressure  exerted  by  this  vapor  is  equal  to  the  vapor  tension 
or  saturated  steam  pressure  in  the  case  of  water,  corresponding 
to  the  temperature  of  the  liquid.  At  this  time  the  space  above 
is  said  to  be  saturated  and  the  weight  of  vapor  per  cubic  foot  is 
the  weight  of  a  cubic  foot  of  saturated  vapor  given  in  the  tables. 
If  the  amount  of  moisture  in  the  air  is  not  equal  to  this,  the  air 
is  not  saturated  with  moisture  and  the  ratio  of  the  weight  of 
moisture  per  cubic  foot  to  the  weight  when  saturated  is  called 
the  relative  humidity. 

>=~:  w 

It  happens  that  the  ratio  of  the  pressure  exerted  by  this 
moisture  compared  with  that  at  saturation  is  practically  the  same, 
since  the  pressure  of  a  perfect  gas  is  proportional  to  the  mass 
when  the  volume  and  temperature  are  constant.  Of  course 
this  is  not  a  perfect  gas  and  the  statement  is  not  absolutely  true. 

.    '    p  =  •'       do) 


CONDENSERS,  COOLING  TOWERS  AND  EVAPORATORS   343 

To  find  the  relative  humidity  a  hygrometer  is  used.  The 
common  form  used  in  engineering  work  is  that  consisting  of  two 
thermometers,  one  of  which  has  its  bulb  surrounded  by  a  damp 
piece  of  wicking.  As  this  is  twirled  at  a  high  rate  in  the  air  the 
evaporation  of  the  water  on  this  bulb  lowers  its  temperature,  the 
amount  of  lowering  being  dependent  on  the  relative  humidity. 
The  fall  of  temperature  is  then  compared  with  the  temperature 
of  the  dry  bulb  and  the  barometric  reading;  and  from  tables 
or  formulae  the  relative  humidity  is  found. 

Carrier  proposes  the  formula 

p'       Bar,  -p'  t-t' 

pt  Pt  2755  -  1.28*' 

p'  =  pressure  corresponding  to  wet  bulb  temperature 
pt  =  pressure  corresponding  to  dry  bulb  temperature 
Bar.  =  barometric  pressure 
t  =  temperature  of  dry  bulb 
tf  =  temperature  of  wet  bulb. 

If  now  the  air  enters  the  cooling  tower  at  a  temperature  fa 
and  of  relative  humidity  pi  the  amount  of  moisture  per  cubic 
foot  is 


where  mi  =  weight  of  1  cu.  ft.  of  steam  at  temperature  fa. 

One  cubic  foot  of  entering  air  is  considered  in  problems  of 
the  cooling  tower  as  it  leads  to  simple  calculations. 

If  the  temperature  of  the  warm  water  is  ti}  the  temperature 
fa  of  the  air  leaving  may  be  taken  as  10  to  20°  below  U  but  satu- 
rated on  account  of  the  intimate  mixture.  The  water  leaving 
the  tower  will  probably  be-above  the  temperature  fa  of  the  en- 
tering air  although  it  might  be  equal  to  it  in  some  cases.  Assume 
fa  to  be  10°  F.  or  20°  F.  below  the  temperature  t0  of  the  outlet 
water.  With  intimate  mixture  fa  may  correspond  to  t0  and  fa 
to  fa. 

The  moisture  per  cubic  foot  leaving  the  tower  is  m2  but  the 

number  of  cubic  feet   have  been   changed  on  account  of  the 

changes  of  temperature  and  pressure.     Let  the  weight  of  air  be  M. 

i,  __  (bar.  -  pipOl447i       (bar.  -  p2)14472 

BTi  BT2 

Bar.  =  barometric  pressure  in  Ibs.  per  sq.  in. 
Pi  =  relative  humidity  at  1 


344  HEAT  ENGINEERING 

Pi  =  saturation  pressure  at  1 
TI  =  absolute  temperature  at  1 
Vi  =  volume  at  1  =  1  cu.  ft. 

The  volume  of  air  at  outlet  corresponding  to  1   cu.  ft.  at 
entrance  is 


The  weight  of  moisture  evaporated  per  cubic  foot  of  original 
air  is 

me  =  m2V2  —  pirai  (14) 


If  Ma  =  weight  of  cooling  water  cared  for  by  1  cu.  ft.  of  air 
at  entrance,  the  weight  of  water  remaining  is 

Ma  -  me 

The  late  Prof.  H.  W.  Spangler  solved  this  problem  by  equating 
energies  at  entrance  and  exit  and  hence  it  is  necessary  to  find 
the  energy  brought  in  by  each  substance  above  32°  F. 

I.  Energy   in  water    at    bottom  of    tower    above  32°  F.  = 

(Ma  -  me)  q'0  (15) 

II.  Energy  in  air  at  entrance  above  32°  F.  = 

A  (bar.  -  pi??i)144  X  1        TT 


III.  Energy  in  moisture  entering  above  32°  F.  = 

piwi  [ii  -  AP&'"]  =  pimiii  -  Api  X  1  (17) 

i  is  heat  content  of  superheated  steam  at  pressure  pipi  and  of 
superheat  of  ti  —  ts  degrees. 

IV.  Energy  in  water  at  top  of  tower  above  32°  F.  =  Maq'i  (18) 

V.  Energy  in  air  at  exit  above  32°  F.  = 

A  (bar.  -  p2)144F2        TJ  nox 

14  _  l  ~  UM 

VI.  Energy  in  moisture  leaving  above  32°  F.  = 

m2V2  [q'z  +  r.2]  -  Ap2V2 
or  better  U  =  ra2F2  [^2  +  p2].  (20) 

VII.  Work  done  by  air  in  changing  1  cu.  ft.  to  V2  cu.  ft.  at 
barometric  pressure  = 

A  bar.  X  144  X  (V2  -  1)  (21) 


CONDENSERS,  COOLING  TOWERS  AND  EVAPORATORS     345 

Now  the  sum  of  the  energies  entering  must  equal  the  sum 
of  those  leaving  plus  the  work: 

II  +  III  +  IV  =  1+  V  +  VI  +  VII  (22) 

In  this  equation  the  only  unknown  is  Ma  and  this  may  be 
found.  From  Ma  the  number  of  cubic  feet  of  air  required  for 
a  given  installation  may  be  known  and  from  this  the  size  of  the 
fan,  power  required,  size  of  tower  and  other  data  may  be 
ascertained. 

This  is  the  best  method  for  solving  such  a  problem  as  other 
methods  of  attack  are  open  to  objections  since  the  temperature 
at  which  events  take  place  is  not  known. 

PROBLEM 

Suppose  air  at  70°  gives  a  wet  bulb  temperature  of  58.5°  F.  and  is  used  in 
a  cooling  tower  with  hot  water  at  140°  F.  The  barometer  is  14.7  Ibs.  The 
air  is  so  mixed  that  it  leaves  at  140°  F.  while  the  leaving  water  is  cooled  to 
80°  F.  How  much  air  would  be  required  to  cool  30,000  Ibs.  of  water  per 
hour. 

P58-5  =  0.2428 

PTO     =  0.3627 

0.2428       14.7  -  0.2428  70  -  58.5 


0.3627  0.3627          *  2755  -  1.28  X  58.5 

=  0.668  -  0.171  =  0.497  =  0.5 
mi  =  0.001152 

=  0.000576 
m2  =  0.00814 

pz  =  2.885 

_  14.7  -  0.5  X  0.3627       601  _ 

14.7  -  2.885          X  531  ~     d9 

m272  =  0.00814  X  1.39  =  0.0113 
me  =  0.0113  -  0.000576  =  0.0107 

Energy  in  water  at  bottom  =  (Ma  -  0.0107)48.1  =  48.1Ma  -  0.515. 
Energy  in  air  at  entrance  =  ==«  - —    — ^-^ —  C732  =  6.70  —  C732. 

Energy  in  moisture  =  0.000576  [19.1  +  1061.8  +  19  X  0.43]  -  [     -  X 

L778 

0.1814  X  1441   =  0.600. 

(Moisture  at  entrance  is  under  a  pressure  of  0.1814  Ib.  and  at  a  temperature 
of  70°.  0.1814  Ib.  means  a  saturation  temperature  of  51°  or  the  moisture  is 
19°  superheated.  From  the  curves  of  specific  heats  in  Chapter  I,  cp  is  0.43.) 


346  HEAT  ENGINEERING 

Energy  in  water  at  top  of  tower  =  Ma  X  108.0. 

_.,  .       .  1    [14.7  -  2.885J144  X  1.4      TT 

Energy  in  air  at  top  =  ^  —  — ^ —  C732  =  7.65  —  t/32. 

Energy  in  moisture  at  top  =  0.0113[108.0  +  947.5]  =  11.9. 

Work  done  by  changing  1  cu.  ft.  of  volume  to  1.4  cu.  ft.  of  volume  at 

atmospheric  pressure  =  ^s  X  14.7  X  144(1.4  -  1)  =  1.09. 

I  I  O 

6.70  -  C732  +  0.600  +  Ma  X  108.0  =  48.1Ma  -  0.515  +  7.65  -  U32 

+  11.9  +  1.09 
1 2  82 

M°  -  56T9-  "  °'214 

This  is  the  amount  of  water  cared  for  by  1  cu.  ft.  of  air  at  entrance,  the 
reciprocal,  4.7  cu.  ft.,  being  the  number  of  cubic  feet  of  air  required  to  care 
for  1  Ib.  of  hot  water.  The  amount  of  water  evaporated  per  cubic  foot 
of  air  has  been  found  to  be  0.0107  Ib.  This  quantity  is  5.2  per  cent,  of 
the  water  supplied  (0.214  Ib.)  per  cubic  foot  of  air. 

In  the  problem  30,000  Ibs.  of  water  are  passed  through  the 
cooling  tower.  This  requires  4.7  X  30,000  or  141,000  cu.  ft.  of 
air  per  hour.  The  evaporation  will  amount  to  5.2  per  cent,  of 
the  weight  of  water  or  1600  Ibs.  of  water  per  hour.  Had  this 
cooling  tower  been  supplied  with  water  at  a  lower  temperature 
the  evaporation  would  have  been  much  less  but  the  amount  of 
air  would  be  greater.  With  the  temperature  of  the  condensing 
water  from  a  condenser  at  140°  F.  (this  temperature  is  much 
higher  than  that  found  in  practice)  with  a  supply  temperature 
of  80°  F.,  16  Ibs.  of  water  would  be  required  per  pound  of 
steam  so  that  30,000  Ibs.  of  water  would  be  used  with  1880  Ibs. 
of  steam.  If  this  condensed  steam  were  used  in  a  cooling  tower 
it  would  more  than  care  for  the  loss  by  evaporation  and  if  used 
for  boiler  feed,  the  make-up  water  to  be  bought  would  be  less  than 
the  feed  water  saved.  In  all  cases  where  the  condensate  can  be 
used  either  for  boiler  feed  or  condensing  water  the  cooling  tower 
reduces  the  bill  for  water.  If  the  water  cannot  be  used  it  will 
be  found  in  most  cases  that  the  sum  of  the  make-up  water  and  feed 
water  will  be  less  than  the  steam  necessary  with  a  non-condensing 
plant.  In  all  cases  the  heating  of  the  air  and  the  doing  of  ex- 
ternal work  makes  the  amount  of  evaporation  in  the  tower  less 
than  the  amount  of  condensation  cared  for  by  the  cooling 
apparatus. 

It  will  be  well  to  keep  in  mind  that  although  the  air  entering 
the  lower  part  of  the  cooling  tower  be  saturated  with  moisture, 
due  to  rain  or  snow,  the  greater  moisture  content  for  the  warm 


CONDENSERS,  COOLING  TOWERS  AND  EVAPORATORS     347 

air  makes  evaporation  possible  even  in  this  case.     The  warming 
of  the  air  also  removes  heat  and  cools  the  water. 


SIZE  OF  TOWER  AND  MATS 

The  size  of  the  tower  necessary  to  be  used  may  be  found  by 
assuming  the  air  to  pass  through  at  a  velocity  of  700  ft.  per  min- 
ute and  the  area  of  the  mats  or  cooling  surface  may  be  found  by 
allowing  200  B.t.u.  per  hour  per  square  foot  of  surface  with  10° 
cooling  to  700  B.t.u.  per  hour  per  square  foot  with  35°  cooling. 
This  may  sometimes  be  expressed  in  terms  of  the  water,  1  sq. 
ft.  for  25  Ibs.  of  water  per  hour.  This  is  independent  of  the 
amount  of  temperature  change. 

For  the  problem  above  the  net  area  of  the  horizontal  cross  section  of  the 
tower  would  be 

Fn  =  60  X700  =  3'36  Sq'  ft' 
The  area  of  the  mats  would  be 

30000(140  -  80)  _ 
Fm  ~          1000*  )0 sq.ft. 

*  This  is  assumed  for  the  large  drop  of  60°  F. 
Fm  =  ?^P  =  1200  sq.  ft. 

In  planning  this  tower  care  should  be  taken  to  prevent  the 
water  from  falling  free.  It  must  be  kept  in  contact  with  the 
mats.  There  must  be  care  in  arranging  the  passages  to  prevent 
air  from  taking  any  path  which  does  not  bring  it  into  contact 
with  the  water. 

Towers  cost  from  $1.00  to  $6.00  per  kilowatt  capacity  or  as 
much  in  some  cases  as  the  condenser  equipment. 

The  power  to  drive  the  fan  is  about  2  to  5  per  cent,  of  the 
engine  power.  In  the  case  above  the  practical  use  of  the  fan  is 
to  give  velocity  only,  as  the  drop  in  pressure  is  practically  nothing. 
For  this  reason  the  formula  for  the  power  of  the  fan  need  only 
consider  the  kinetic  energy  of  the  air,  the  friction  head  being 
considered  as  equal  to  the  velocity.  This  gives 

141, 000  X  14.7  X  144  /700\ 
work  per  minute  =  2  X  -^TT^^^F^r^i hsr) 


60  X  53.35  X  531        \  60  /    64.4 
=  750  ft. -Ibs.  per  min. 


348  HEAT  ENGINEERING 

SPRAY  NOZZLES  AND  PONDS 

Another  device  often  used  to  cool  water  is  the  spray  foun- 
tain. Nozzles  are  supplied  with  hot  water  from  a  main  and  the 
velocity  of  discharge  is  made  to  divide  it  into  a  fine  spray.  The 
water  is  then  caught  in  a  reservoir  from  which  it  is  taken  to  the 
condenser.  With  these  nozzles  the  water  may  be  cooled  15° 
when  the  atmosphere  is  25°  to  30°  below  the  temperature  of  the 
hot  water  while  20°  may  be  obtained  with  a  40°  difference.  These 
nozzles  are  placed  about  8  ft.  apart  and  are  usually  fitted  to  3-in. 
pipes  in  which  the  velocity  should  be  about  5  ft.  per  second. 
In  this  way  each  nozzle  will  care  for  about  60,000  Ibs.  of  water 
per  hour. 

In  some  plants  large  cooling  ponds  are  used  to  cool  the 
water  by  surface  evaporation.  With  these  the  hot  water  is 
discharged  at  one  end  of  the  pond  and  the  cooling  water  is  taken 
off  at  the  other.  Thomas  Box,  many  years  ago,  recommended 
that  210  sq.  ft.  of  cooling  surface  be  used  per  nominal  horse- 
power if  the  engine  was  operated  for  24  hr.  per  day.  This 
was  with  engines  using  more  steam  than  those  of  to-day  and 
this  might  be  reduced  to  say  120  sq.  ft.  per  24  h.p.-hr. 
per  day,  or  about  5  sq.  ft.  per  horse-power  hour  during  the  day 
of  24  hr.  W.  B.  Ruggles,  in  the  Journal  of  the  A.S.M.E. 
for  April,  1912,  gave  a  test  of  a  cooling  pond  of  288,000  sq.  ft.  of 
a  depth  of  5.38  ft.  in  which  he  found  a  transfer  of  3.67  B.t.u. 
per  square  foot  of  water  surface  per  hour  per  degree  difference  in 
temperature  between  air  and  water,  and  that  120  sq.  ft.  of  surface 
per  horse-power  was  sufficient  when  16  Ibs.  of  steam  were  used 
per  horse-power  hour. 

ACCUMULATORS  AND  EVAPORATORS 

Regenerator  accumulators  are  devices  used  for  the  retention 
of  surplus  heat  until  needed  when  •  an  intermittent  supply  is 
given  off  by  a  machine  and  it  is  desired  to  use  this  in  another  piece 
of  apparatus.  They  were  planned  by  Rateau  to  be  used  with 
low-pressure  turbines  which  were  operated  by  the  exhaust 
steam  from  engines,  the  operation  of  which  was  intermittent  as 
is  the  case  with  rolling  mill  engines  or  from  steam  hammers. 
One  of  his  forms  is  shown  in  Fig.  167.  It  consists  of  a  vessel  A 
made  of  steel  in  which  there  is  a  horizontal  partition  at  the  center. 


CONDENSERS,  COOLING  TOWERS  AND  EVAPORATORS     349 

A  number  of  elliptical  flues  B.B.  are  placed  in  the  tank.  These 
receive  steam  from  the  pipe  D  and  manifold  C  and  steam  is  dis- 
charged through  %-in.  holes  in  the  flue  walls  into  the  water  carried 
in  the  chambers.  This  water  may  enter  the  flues.  The  steam 
heats  the  water  which  is  introduced  from  the  float  box  E  as  soon 
as  the  water  level  is  lowered.  The  steam  entering  from  D  may 
also  enter  the  top  of  the  accumulator  by  the  check  G  and  pass 
out  to  the  turbine  through  F.  If  however  the  intermittent  supply 
is  reduced  or  stopped  momentarily,  the  pressure  in  the  accu- 
mulator falls  and  the  warm  water  begins  to  evaporate,  thus  main- 
taining the  steam  supply.  Of  course  if  the  supply  of  steam  is 
discontinued  for  a  long  time,  the  turbine  receives  its  steam 
through  a  reducing  pressure  valve  from  the  boiler  main.  Baffle 
plates  are  used  above  the  elliptical  steam  flues  to  cause  the  water 
and  steam  to  mix  and  to  dry  out  the  steam  leaving  the  accu- 


A 

1 

B- 

O      O       0 

000 

000 

000 

0  0     0      1 

n  1 

FIG.  167. — Rateau  accumulator. 

mulator.  In  some  Rateau  Accumulators  large  masses  of  iron 
in  the  form  of  trays  are  used  to  absorb  the  heat  and  supply  that 
necessary  to  evaporate  the  water  when  the  supply  is  reduced. 

Evaporators  are  used  for  the  purpose  of  concentrating  liquors  or 
solutions  and  for  the  production  of  distilled  water  or  other  liquid 
although  when  used  for  this  latter  purpose  they  are  known  as 
stills.  Fig.  168  shows  one  form  of  evaporator.  In  this  one  a 
liquid  to  be  evaporated  is  carried  to  the  line  A  in  a  tank  B  con- 
taining a  set  of  tubes  C  held  between  two  tube  plates.  The  space 
D  between  the  tube  plates  is  separated  from  the  remainder  of  the 
shell.  This  space  is  supplied  with  steam  or  some  other  hot  vapor. 
The  vapor  gives  up  its  heat  to  the  fluid  within  the  tubes  and  is 
condensed,  the  condensate  leaving  at  E.  If  the  pressure  in  the 
chamber  above  the  level  of  the  liquid  is  such  that  the  boiling 
temperature  of  the  liquid  is  below  the  temperature  of  the  vapor 
entering  the  space  D  from  the  pipe  F,  the  liquid  will  boil  and  its 


350 


HEAT  ENGINEERING 


vapor  may  be  used  to  boil  liquid  in  a  second  chamber  in  which 
the  pressure  is  maintained  still  lower  than  that  in  the  first  cham- 
ber. This  operation  may  be  repeated  as  shown  in  the  figure, 
the  vapor  from  the  last  chamber  B"  passing  to  the  condenser  H 
in  which  the  pressure  is  maintained  at  a  low  point  by  the  air  pump 
/.  In  many  cases  where  these  are  used  for  the  concentration  of 
liquors  as  in  sugar  making,  the  dilute  solution  enters  at  J  and  as 
it  becomes  more  concentrated  it  sinks  to  the  bottom  of  B  and  is 
passed  over  to  the  uppfer  level  of  Kf  of  the  next  evaporator. 
Since  the  pressure  is  lower  in  C  than  in  B  this  action  will  take  place 
and  be  regulated  by  opening  a  valve  in  the  line.  This  action  is 
carried  on  throughout  the  system. 


FIG.  168. — Triple  effect  evaporator. 

Where  three  of  these  are  used  as  shown  in  Fig.  168,  the  arrange- 
ment is  known  as  a  triple  effect  evaporator,  a  single  one  as  a 
single  effect,  two  as  a  double  effect,  four  as  a  quadruple  effect. 

Assuming  the  arrangement  as  shown,  suppose  M  pounds  of 
liquor  enter  at  J.  Of  this  e\  pounds  are  evaporated  and  M  —  ei 
pounds  pass  on  to  the  second  effect  to  be  evaporated.  Of  this 
ez  are  evaporated  and  M  —  (61  +  62)  are  passed  on  and  63  are 
evaporated  and  M  —  (61  +  62  +  63)  are  drawn  off  as  a  con- 
centrated solution. 

In  the  first  effect  suppose  1  Ib.  of  vapor  is  admitted  at  F. 
This  is  condensed  and  the  condensate  is  removed  at  E.  This 
weighs  1  Ib.  61  pounds  of  vapor  are  admitted  to  Ff  and  61 
pounds  of  vapor  are  condensed  here.  62  pounds  are  introduced 
and  condensed  in  the  third  stage. 

In  working  out  the  condensation  and  evaporation  in  the  differ- 


CONDENSERS,  COOLING  TOWERS  AND  EVAPORATORS     351 

ent  effects  the  method  of  procedure  is  to  start  with  the  one  effect 
and  work  to  the  others.  In  these  various  tanks  the  amount  of 
heat  loss  must  be  allowed  for  and  if  the  size  of  the  tanks  be 
assumed  the  amount  of  heat  loss  per  hour  may  be  computed  for 
each  from  the  value  K  for  coverings  as  given  in  Chapter  III. 
The  pressures  in  the  various  tanks  are  assumed  so  as  to  give  the 
proper  temperature  differences.  These  may  be  made  20°  for 
each  stage. 

The  condensates  from  the  various  stills  could  be  discharged 
through  a  feed  heater  and  this  heat  could  be  added  to  the  feed 
entering  the  first  effect  so  that  the  condensates  would  leave  at 
the  temperature  of  the  fresh  liquid.  The  same  could  be  done  in 
theory  with  the  condensate  from  the  condenser.  The  operation 
of  the  triple  effect  depends  on  a  difference  in  the  temperatures  of 
the  various  effects  and  hence  in  any  problem  the  initial  steam 
pressure  pi  and  temperature  t\  and  the  final  temperature  tc 
would  be  assumed  and  with  them  the  various  intermediate  tem- 
peratures, from  the  temperatures  of  the  condenser  and  the  tem- 
perature of  the  condensate.  For  a  triple  effect,  the  temperature 
of  the  steam  in  the  first  effect  is  ti  in  the  supply  and  £2  in  the  dis- 
charge while  £3  is  the  temperature  of  the  discharge  from  the 
second  and  tc  is  the  temperature  of  the  discharge  from  the  last 
stage.  The  temperature  of  the  cooling  water  and  weak  liquor 
is  t{  and  that  of  the  warm  circulating  water  is  t0  while  the  con- 
densate in  the  condenser  is  fo.  The  following  conditions  hold  for 
the  various  stages  if  10  per  cent,  of  the  heat  entering  is  assumed 
to  be  lost  in  radiation. 

First  Stage — Heat  Entering: 

With  1  Ib.  steam  Q  =  q\  +  n  (23) 

With  ei  +  €2  +  €3  Ibs.  feed  Q  =   (ei  +  e%  +  e^q'i 
+  Ifa'i  -  <?'<)  +  eitfa  -  <?';)  +  e2(q's  -  <?'<)  + 

ez(qfh  -  <?';).  (24) 

(Feed  heaters  are  used  to  reduce  condensates  to  fc) 
Heat  Leaving : 

With  ei  Ibs.  evaporation  Q  =  ei(q'z  +  r2)  (25) 

With  62  +  £3  Ibs.  feed  to  next  evaporator  Q  =  (e2  +  e^q'z  (26) 

With  radiation  Q  =  Kofe'i  +  n)  (27) 

With  1  Ib.  condensate  Q  =  q\  (28) 


352  HEAT  ENGINEERING 

The  equation  for  heat  balance  is 


r2)  +  (02 

0.9(^1  +  n)  -  <?'t-  =   6ir2  +  e2(g'2  -  ?'3)  +  e,(g'2  -  tf'/O      (29) 
The  equation  for  the  second  stage  in  the  same  manner  is 

0  =  ei[0.9r2  -  O.ltf's]  -  e2[?'3  +  r3  -  g'J  -  e3[g'3  -  ff'2]   (30) 
The  third  stage  gives 

0  =  62[0.9r3  -  0.1  g'3]  -  ez[q'c  +  rc  -  q'3]  -  m0q'c  (31) 

These  three  equations  are  sufficient  to  find  the  quantities 
€1,  62  and  63. 

In  the  last  effect  the  final  withdrawal  of  the  concentrated 
liquor  will  give  the  subtractive  term  in  the  last  equation: 

m0q'c  (32) 

m0  is  very  small  compared  with  the  other  terms. 

PROBLEM 

As  a  problem,  suppose  water  at  65°  F.  is  to  be  distilled  in  a  triple  effect 
with  steam  at  228°  F.  with  10  per  cent.  loss.  Suppose  the  temperatures  of 
boiling  are  208°,  188°  and  168°  and  that  the  temperature  of  the  outlet  water 
from  the  condenser  is  85°  and  the  hot  well  is  90°  F.  The  equations  above 
then  become 

0.9  [196.5  +  959.4]  -  33.1  =  ei(972.2)  +  e2(176.2  -  156.1) 

+  e3(176.2  -  58.1) 

0  =  6l[0.9  X  972.2  -  0.1  X  176.2]  -  e2[156.1  +  984.7  -  176.2] 

+  e3  [176.2  -  156.1] 

0  =  e2[0.9  X  984.7  -  0.1  X  156.1]  -  e3[136.0  +  996.7  -  156.1] 

1007.3    =  972.26l  +  20.1e2  +  118.1e3 
0    =  857et  -  964.6e2  +  20.1e3 
0    =  870.7e2  -  976.6e3 
e2  =  1.12e3  =  0.84 
ei  =  1.21e3  =  0.93 
es  =  0.75 

The  weight  of  condensate  =  3.52  Ibs. 

The  amount  of  water  used  in  the  condenser  per  pound  of  steam  in  the 
first  evaporation  is 

M  =    0-75  [136.0  +  996.7  -58.11    _  ^  ^ 
20 

The  amounts  of  heat  can  now  be  computed  for  each  evaporator  when  the 

M 
total  amount  of  evaporation  is  known.     If  this  amount  is  M,  *s  ^e 


CONDENSERS,  COOLING  TOWERS  AND  EVAPORATORS     353 


amount  of  steam  condensed  in  the  first  stage, 
second  stage,     T    M  is  used  on  the  third  and 


0  93 


is  condensed  in  the 

condensed  in  the  con- 

denser.    By  using  the  methods  of  this  chapter  or  Chapter  III,  the  surface 
required  for  this  heat  under  the  conditions  given  can  be  determined. 

The  loss  of  heat  from  the  surface  of  the  evaporators  can  be 
computed  by  the  methods  of  Chapter  III,  when  the  actual  size 
and  shape  is  known  with  the  temperatures.  This  would  give 
a  definite  number  of  heat  units  instead  of  the  percentage.  Its 
use  would  be  the  same  as  above. 

At  times  the  supply  for  each  evaporator  is  taken  from  the 
weak  liquor  line  and  then  the  terms 


will  be  omitted  from  the  equation.  Each  stage  will  have  the 
weight  evaporated  in  that  stage  as  the  unknown  in  the  equation. 
The  feed  might  be  heated  by  the  outgoing  condensate  or  if  not 
an  assumption  of  its  temperature  would  be  made. 


100.4Tlbu. 


492.00  Ibs.^  83.97  Ibs.  ,*,  76.49  Ibs.  fa  69.31  Ibs.  ^  €2.73  Ibg. , 

Ii 
|A- 


Condenser 
Ib. 


SupplyQ 


FIG.   169. — Hodge's  multiple  still. 


To  remove  the  air  which  may  collect  in  the  various  condensing 
chambers,  air  pumps  must  be  attached  or  small  vent  holes 
must  be  made  from  the  condensing  chamber  to  the  boiling 
chamber. 

Another  multiple  effect  known  as  Hodge's  Multiple  Still  is 
shown  in  Fig.  169.  In  this  A  is  the  boiler  and  B,  C,  D,  E,  F,  G 
and  H  are  evaporators  or  stills.  Steam  from  the  boiler  is 
passed  into  still  B  to  evaporate  its  amount  M  of  the  liquid 
and  enough  steam  is  added  to  the  steam  exhausted  from  one 
still  from  the  boiler  steam  to  give  sufficient  heat  to  condense 
the  same  quantity  M  in  the  next  still  and  to  make  up 
for  the  radiation  and  the  steam  lost  by  the  blow-off  from 
the  valves  K.  The  blow-off  is  intended  to  remove  the  air  from 

23 


354  HEAT  ENGINEERING 

the  evaporator.  /  is  a  condenser  and  J  is  a  feed-water  heater 
to  reduce  the  temperature  of  the  condensate.  The  feed  to  each 
still  is  controlled  by  a  float  valve  and  this  feed  together  with 
the  boiler  feed  is  carried  through  the  same  line.  Since  the 
water  used  in  the  condenser  is  sufficient  to  feed  the  boiler  and 
stills,  its  amount  is  limited.  Since  in  addition  to  this  the  maxi- 
mum possible  temperature  leaving  the  condenser  is  fixed  by  the 
pressure  of  the  steam  to  be  condensed,  the  temperature  entering 
the  condenser  is  fixed  and  hence  this  water  might  not  be  suffi- 
cient to  cool  the  condensate  in  the  feed-water  heater  to  a  low 
value.  The  extra  cooling  water  is  required  to  reduce  the  tem- 
perature of  the  condensate  still  lower.  Although  not  used  in 
the  Hodge's  Multiple  Still,  a  feed  heater  on  the  outlet  of  the 
condensate  from  each  still  would  increase  the  efficiency  if  the 
feed  to  that  still  were  passed  through  it. 

To  design  such  a  still  the  desired  amount  of  distillation,  or 
the  capacity,  and  the  pressures  and  temperatures  would  be 
assumed.  The  total  capacity  would  be  divided  equally  among 
the  stills  and  the  condenser;  in  the  above  figure  among  eight  units. 
From  the  amount  of  water  to  be  evaporated  or  condensed,  to- 
gether with  the  temperature  difference,  the  probable  size  of  each 
unit  is  found.  From  the  size  and  temperature  the  amount  of 
heat  lost  may  be  computed.  The  problems  are  now  solvable 
since  equations  for  each  still  may  be  written  out  as  soon  as  the 
amount  of  blow  is  assumed. 

PROBLEM 

Suppose  400  Ibs.  of  distilled  water  are  desired  per  hour  and  from  the  size 
of  still  required  to  condense  #  X  400  Ib.  or  50  Ibs.  the  radiation  is  4000 
B.t.u.  per  hour  and  that  %  Ib.  of  steam  blown  per  hour  would  "be  sufficient 
to  keep  the  evaporator  clear  of  air. 

The  temperature  drop  in  each  unit  jvill  be  taken  as  10°  and  since  the 
pressure  even  in  the  condenser  is  to  be  above  the  atmosphere  its  temperature 
must  be  above  212°  F.  Suppose  this  is  taken  as  220°  F.,  then  the  tempera- 
tures of  the  various  stills  will  be  220°,  230°,  240°,  250°,  260°,  270°  and  280°. 
The  temperature  of  the  boiler  steam  will  be  290°  F.  The  absolute  pressure 
will  therefore  be  17.2  Ibs.,  20.8  Ibs.,  25.0  Ibs.,  29.8  Ibs.,  35.4  Ibs.,  41.8  Ibs., 
49.2  Ibs.  and  57.5  Ibs. 

The  amount  of  steam  and  water  leaving  the  apparatus  is 

8  X  50  +  8  X  %  =  406  Ibs. 

This  must  be  the  total  feed. 

The  temperature  of  the  condensing  water  leaving  the  condenser  would  be 


CONDENSERS,  COOLING  TOWERS  AND  EVAPORATORS     355 

210°  F.  if  the  same  temperature  difference  is  to  be  used  here.     The  feed  water 
entering  any  still  is  at  this  temperature. 
The  equation  for  the  condenser  is 


+  503/4(g'220  -f  r220)  =  K(q'MO  +  r220)  +  50g'220  +  406g'210 

+  4000    (33) 
q'i  =  69.0 
U  =  101° 
The  equation  for  H  is 

(50%  +  Mb)(q'z30  +  r230)  +  M/g'210  =  50g'230  +  50%(g'220  +  r220) 

+  %te'23o  +  r230)  +  4000    (34) 

50%  +  Mb  +  M/  =  50%  +  50  +  %  (35) 

M&  +  M/  =  50%  (36) 

Mb  =  5.79  (37) 

M/  =  44.96  (38) 

The  equation  for  G  will  be 

(56.54  +  M6)(g'24o  +  r240)  +  Mfq'2lo  =  50g'24o  +  56.54(g'a»0  +  r230) 

+  %  (^240  +  ^40)  +4000     (39) 

56.54  +  Mb  +  Mf  =  50  +  56.54  +  H 
Mb  +  Mf  =  50% 
Mb=  6.41 
Mf  =  44.34 

The  amounts  for  the  stages  are  as  follows: 

Mb  Mf 

Stage  5  8.44  42.31 

Stage  C  8.03  43.00 

Stage  D  7.48  42.98 

Staged  7.18  43.48 

Stage  F  6.58  44.11 

Stage  G  6.19  44.70 

Stage  #  5.79  44.96 

Condenser  50.75 


Feed  to    boiler      100.44  Feed  to  tank  305 . 56 

Total  feed  =  406  Ibs. 

406 
The  amount  of  evaporation  per  pound  of  steam  is  1QQ  .^  =  4.05.     A 

result  that  is  only  a  little  better  than  could  be  obtained  with  an  ordinary 
quintuple  effect. 

The  temperature  of  the  distillate  is  found  by  adding  together  the  heats  of 
the  liquid  for  each  discharge  and  then  dividing  this  by  eight  to  get  the  aver- 
age heat  of  the  liquid.  This  is  true  because  each  evaporator  sends  50  Ibs. 
into  the  distillate  line.  This  gives 

qrm  =  223.9 
tm  =  255°.l 


356  HEAT  ENGINEERING 

The  temperature  of  the  water  entering  the  condenser  will  have  to  be  101° 
so  that  the  heat  necessary  to  raise  this  water  from  65°  F.  will  reduce  the  tem- 
perature of  the  distillate 

406(101  -  65) 


400 


36.3°  F. 


or  to  218.8°  F.  To  cool  this  off  still  more  cold  water  is  circulated  in  the 
cooler.  If  the  distillate  is  desired  to  be  cooled  to  80°  F.  the  amount  of 
water  needed  would  be 

400(218.8-80] 
M  =         80-  65 

This  heat  could  be  saved  by  arranging  the  distillate  to  warm  the  feed  at 
each  still. 

DOUBLE  BOTTOMS 

The  design  of  double  bottoms  for  evaporation  of  liquors  is 
carried  out  in  the  manner  of  Chapter  III.  The  temperature  of 
boiling  of  the  liquor  may  be  fixed  by  the  pressure  and  found  from 
a  table  or  if  the  temperature  is  given  the  pressure  may  be  found 
in  tables.  If  the  tables  are  not  at  hand  recourse  may  be  had  to 
the  rule  of  Diihring  as  given  by  Hausbrand: 


TEMPERATURE  OF  BOILING 

The  difference  between  the  boiling  temperatures  of  a  liquid  at 
any  two  pressures  is  equal  to  a  constant  multiplied  by  the  dif- 
ference in  temperatures  of  water  at  the  same  pressures.  The 
values  of  the  constants  are  given  for  some  substances  in  the 
table  below: 

Water  1 . 0 

Alcohol  0.904 

Ether  1 . 0 

Acetic  acid  1 . 164 

Benzene  1 . 125 

Turpentine  1 . 329 

Mercury  2 . 0 

Carbolic  acid  1.20 

If  one  temperature  and  pressure  is  known  for  a  substance 
another  may  be  found.  Thus  if  mercury  boils  at  674°  F.  at  at- 
mospheric pressure  it  is  desired  to  find  at  what  pressure  it  would 
boil  at  400°  F.  Water  at  atmospheric  pressure  boils  at  212°  F. 
The  temperature  for  water  corresponding  to  mercury  at  400°  F. 
is  given  by; 

674  -  400  =  2  (212  -  tx) 


CONDENSERS,  COOLING  TOWERS  AND  EVAPORATORS    357 

Since  the  constant  in  the  table  is  2. 

tx  =  75°. 
Pressure  =  0.42  Ibs.  per  sq.  in 

Having  the  temperature  required,  the  temperature  difference 
may  be  found  or  assumed  and  then  if  the  heat  of  vaporization 
is  found  by  reference  to  tables  the  area  may  be  found  by 


. 

Where  k  =  1600  for  water. 

k  =  1200  for  thin  liquors. 

k  =  500-900  for  thick  liquors. 

TOPICS 

Topic  1.  —  What  is  the  purpose  of  the  dry  air  pump?  Why  is  the  air  for 
this  pump  brought  into  contact  with  the  coldest  water?  Explain  clearly. 
On  what  does  the  amount  of  air  depend  ?  What  are  the  peculiar  features  of 
the  Wheeler  dry  tube  condenser,  the  uniflux  and  the  contraflo  condenser? 
What  is  the  reason  for  the  use  of  the  barometric  condenser? 

Topic  2.  —  Outline  the  method  of  design  for  all  parts  of  a  condenser. 

Topic  3.  —  Sketch  and  explain  action  of  a  cooling  tower  and  derive  equation 
for  the  determination  of  the  amount  of  water  cooled  per  cubic  foot  of  air 
taken  in.  Give  the  expressions  for  computing  each  term  of  the  equation. 

Topic  4.  —  Explain  the  action  of  spray  nozzles,  ponds,  and  accumulators 
and  show  how  to  find  size  of  each  to  perform  a  given  service. 

Topic  5.  —  Sketch  and  explain  action  of  a  triple  effect.  Write  formulae  for 
the  determination  of  the  evaporation  per  pound  of  steam  supplied  to  the 
first  effect. 

Topic  6.  —  Sketch  and  explain  action  of  Hodge's  multiple  still.  Write 
equations  for  the  condenser  and  the  next  two  stills  by  which  the  quantity  of 
steam  may  be  found. 

Topic  7.  —  Explain  how  to  find  the  temperature  at  which  a  substance  will 
boil  under  a  certain  pressure  if  the  temperature  is  known  for  a  given  pressure. 
What  are  double  bottoms?  How  are  they  designed? 

PROBLEMS 

Problem  1.  —  An  air  pump  is  used  with  a  10,000-kw.  turbine.  The  tem- 
perature of  the  hot  well  is  65°  F.  The  vacuum  carried  is  29  in.  How  much 
air  is  present?  If  the  air  and  vapor  are  carried  around  cold  water  pipes  so 
that  the  temperature  is  reduced  to  50°  F.,  how  much  has  the  volume  of  air 
been  reduced?  Is  the  air  leakage  in  this  condenser  system  excessive? 

Problem  2.  —  Find  the  amount  of  surface  to  use  with  a  10,000-kw.  turbine 
with  a  steam  consumption  of  13  Ibs.  of  steam  with  a  vacuum  of  29  in.  and 
temperature  of  steam  of  65°  F.  if  the  cooling  water  operates  from  45°  F.  to 


358  HEAT  ENGINEERING 

60°  F.  How  much  water  is  required?  How  many  H-m.  tubes  would  be 
used  in  a  nest? 

Problem  3. — Find  the  amount  of  air  to  cool  3,900,000  Ibs.  of  water  per  hour 
from  110°  F.  to  80°  F.  in  70°  F..weather  with  the  barometer  at  29.6  in.  and 
the  wet  bulb  at' 60°  F. 

Problem  4. — How  many  spray  nozzles  would  be  required  for  Problem  3? 
How  large  a  cooling  pond  would  be  required? 

Problem  5. — Compute  the  amount  of  distilled  water  made  per  pound  of 
steam  from  a  triple  effect  heated  by  exhaust  steam  at  3  Ibs.  gauge  pressure  if 
the  condensed  steam  in  the  first  effect  is  of  no  value,  x  of  the  entering  ex- 
haust steam  is  0.90.  Temperature  of  water  supply  is  60°  F.  Vacuum 
allowed  on  condenser,  20  in. 

Problem  6. — Find  the  pressure  at  which  mercury  will  boil  at  300°  F. 

Find  the  area  required  for  a  double  bottom  to  drive  off  700  Ibs.  of  water  per 
hour  from  a  solution  at  15  in.  vacuum.  Neglect  heat  of  solution. 


CHAPTER  IX 
INTERNAL  COMBUSTION  ENGINES  AND  COMBUSTION 

The  internal  combustion  engine  has  been  greatly  improved 
within  the  last  25  years  although  its  history  extends  back 
several  centuries.  In  1680  Huygens,  a  Dutch  physicist,  pro- 
posed to  use  the  explosion  of  gun  powder  to  drive  the  piston 
and  in  1690  Papin  continued  this  work.  There  is  nothing  definite 
known  of  their  work  or  similar  proposals  by  Abbe  Hautefeuille 
at  about  this  date.  In  1794  a  patent  was  granted  to  Robert 
Street  for  a  gas  engine  using  turpentine  in  the  bottom  of  a 
heated  cylinder  and  the  mixture  of  this  with  air  was  ignited  by 
a  flame.  Samuel  Brown  originated  an  engine  in  1823—6  in  which 
a  soluble  gas  behind  a  piston  was  dissolved  in  water  and  thus 
produced  a  vacuum  sucking  the  piston  downward.  The  Wright 
engine  in  1833  exploded  a  mixture  of  compressed  gas  and  air 
which  was  admitted  to  the  cylinder  after  the  exploded  gases 
were  driven  out.  This  engine  required  two  revolutions,  or 
four  strokes  to  the  cycle.  The  Barnett  engine  of  1837  used  a 
separate  pump  to  compress  the  air.  In  the  years  1838  to  1854 
there  were  eleven  English  patents  for  gas  engines  applied  for. 
In  1855  the  method  of  igniting  the  charge  by  compressing  it 
into  a  heated  tube  was  patented  by  A.  V.  Newton  and  in  1857 
Barsanti  and  Matteucci  invented  an  engine  with  a  free  piston 
in  which  the  explosion  of  gas  drove  the  piston  upward  and  the 
contraction  due  to  cooling  produced  a  reduction  of  pressure  so 
that  the  piston  was  drawn  back  by  this  and  its  weight,  and 
acted  on  the  crank  shaft  through  a  ratchet. 

The  history  of  successful  gas  engines  begins  with  1860  when 
Lenoir  built  his  machine.  This  resembled  a  steam  engine  in 
structure  and  action.  A  mixture  of  gas  and  air  was  drawn  into 
a  cylinder  by  the  movement  of  a  piston  from  the  fly  wheel  and 
after  the  piston  had  reached  the  middle  of  its  stroke  the  gas 
and  air  were  shut  off  and  the  mixture  was  exploded  by  a  high- 
tension  spark  between  two  platinum  points.  The  pressure 
immediately  rose  and  as  the  piston  moved  forward  this  gas  ex- 
panded reaching  atmospheric  pressure  at  the  end  of  the  stroke. 
On  the  return  stroke,  the  same  thing  occurred  on  the  other  side 

359 


360 


HEAT  ENGINEERING 


of  the  piston,  the  exploded  gases  on  the  first  side  being  ex- 
hausted. The  cycle  of  this  engine  is  shown  by  the  card  of 
Fig.  170. 

In  1863  M.  Alph.  Beau  de  Rochas  published  a  pamphlet  in 
which  he  suggested  that  the  greatest  economy  of  the  gas  engine 
would  be  obtained  if: 

1st.  The  greatest  possible  cylinder  volume  with  the  least 
possible  surface  be  used. 

2nd.     If  the  expansion  be  as  rapid  as  possible. 
3rd.     If  the  expansion  be  as  complete  as  possible. 
4th.     If  the  explosion  pressure  be  as  large  as  possible. 
He  then  follows  this  by  a  description  of  a  cycle  to  give  these 
results.     The  cycle  consists  of: 
Suction  on  entire  stroke. 
Compression  on  second  or  return  stroke. 
Ignition  at  dead  point  and  expansion  on  third  stroke. 
Exhaust  of  gases  on  fourth  stroke. 


1st. 
2nd. 
3rd. 
4th. 


FIG.  170. — Card  of  the  Lenoir 
cycle. 


FIG.  171.— Card  of  the  Beau  de 
Rochas  or  Otto  cycle. 


This  cycle  was  to  have  ignition  due  to  high  compression. 

In  1867  Otto  and  Langen  exhibited  their  first  engine  at  the 
Paris  Exposition.  This  was  similar  to  that  of  Barsanti  but 
it  was  a  practical  machine  although  very  noisy.  While  Lenoir's 
engine  took  over  90  cu.  ft.  of  gas  per  boiler  brake-horse-power 
hour  this  engine  consumed  only  one-half  as  much. 

OTTO  CYCLE 

This  was  followed  by  a  number  of  engines  by  Brayton,  Gilles, 
Halliwell,  Bisschop,  Andrew,  Clerk  and  others,  but  in  1876  Otto 
brought  out  the  Otto  Silent  Engine.  This  engine  operated  on 
the  Beau  de  Rochas  cycle  although  independently  invented  by 
Otto.  This  engine  was  a  great  success  and  the  form  of  cycle 
for  many  years  was  known  as  the  Otto  cycle.  It  is  one  of  the 
commonest  forms  of  cycles.  It  consists,  as  shown  in  Fig.  171, 


INTERNAL  COMBUSTION  ENGINES  361 

of  four  strokes:  First,  a  suction  stroke  ab,  at  a  pressure  slightly 
below  the  atmosphere  due  to  suction  of  the  air  from  the  out- 
side; second,  a  compression  stroke  be  with  explosion  at  the  end 
of  the  stroke,  bringing  the  pressure  from  c  to  d;  third,  an  ex- 
pansion stroke  de  followed  by  free  expansion  ef  to  a  point  just 
above  the  atmosphere;  and  fourth,  a  discharge  stroke  fa  at  a 
pressure  above  the  atmosphere  due  to  the  discharge  being  driven 
out  against  atmospheric  pressure. 

In  the  Beau  de  Rochas  or  Otto  cycle,  the  lines  of  compression 
and  expansion  are  practically  adiabatics  because  the  action  is 
rapid  and  also  because  gas  is  a  poor  conductor.  The  gas  near  the 
cylinder  walls  is  cooled  by  the  water  jacket  used  but  this  transfer 
of  heat  probably  extends  only  to  this  film  of  gas.  The  cycle 
requires  four  strokes  for  its  completion  and  engines  using  this 
form  are  spoken  of  as  four  cycle  en- 
gines. The  cycle  is  practically 
used  on  two  cycle  engines  in  which 
compressed  air  is  admitted  near  e 
driving  out  the  burned  gases  and 
this  is  followed  by  compressed  gas 
so  that  when  the  point  /  has  been 


passed    by  a   small  distance,    the  p^    172.— Otto  cycle, 

burned    gases    have    been    driven 

out  and  the  air  and  its  fuel  gas  have  been  introduced.     In  this 
way  there  is  an  explosion  for  each  revolution. 

To  study  the  cycle  the  lines  fa  and  ab  are  assumed  to  coincide 
and  eliminate  each  other  giving  Fig.  172  as  the  theoretical  cycle. 
This  cycle  is  made  up  of  two  adiabatics  and  two  constant  vol- 
ume lines. 

The  mixture  of  unburned  gases  on  be  and  the  mixture  of  burned 
gases  on  de  are  different  in  character  and  contain  different 
amounts  of  water  vapor  and  carbon  dioxide.  For  this  reason,  the 
lines  are  different  as  carbon  dioxide  and  steam  are  not  perfect 
gases  and  the  variation  from  the  adiabatic  of  a  perfect  gas  is 
different  for  the  two  lines.  Moreover  the  variations  of  the 
specific  heat  as  the  temperature  rises  also  causes  changes  in 
these  lines. 

For  simplifying  computations  so  as  to  study  the  effects  of 
changes  on  the  cycle,  a  preliminary  discussion  is  often  made 
considering  the  gases  present  on  all  four  lines  to  be  air  and 


362  HEAT  ENGINEERING 

considering   the   specific    heats   as    constant    quantities.     Such 
discussions  and  results  are  known  as  results  of  the  air  standard. 


AIR  STANDARD  EFFICIENCY 

Under  these  conditions  and  assuming  1  Ib.  of  substance 
present  on  the  cycle,  the  heats  on  the  various  lines  are  as 
follows : 

Heat  on  be  =  0 

Heat  on  cd  =  cv(Td  -  Tc). 

Heat  on  de  =  0 

Heat  on  eb  =  -  cv(Te  -  Tb) 

Work  =  cv(Td  -  Tc)  -  cv(Te  -  Tb) 

*   (rp  .          rp  \  „   (  rp  rp  \ 

Thermal  eff.  =  r>3  = 


^                     c  (T,  -  T} 

Lv\  J-  d.           J.  c) 
Te   ~    Tb                        Tb                        Te 

w 

(2) 
(3) 

Td- 

Tc-  Tb 

Td 

'     T  ~   x      "    T 
1  c                       id 

~    Te 

Tc 

Te-  Tb 

Tb 

Td 

Te 

Td-  Tc~ 

Tc~ 

~  Td 

Since 

This  is  reduced  from  the  cross  products  of  temperature. 

TeTc  =  TbTd 

This  states  that  the  theoretical  efficiency  of  the  Otto  cycle, 
air  standard,  is  equal  to  the  range  of  temperature  on  either 
adiabatic  divided  by  the  higher  temperature  on  the  adiabatic. 

Tt       /  V  \  k~l 

Now  -,  -  (4) 


But  Vc  is  the  clearance  volume  and  Vi  is  equal  to  the  clear- 
ance volume  plus  the  displacement,  hence 

Vc  ID  I 


Vb       (l  +  l)D       I  +  1 

*-i  i       k-i 


Hence  r;3  =  1  -   x  7   ,    ,  ,          -  .       ,          ^  . 

H 


1  -  / - 

V* 


As  I  decreases  the  last  term  becomes  smaller  and  the  efficiency 
increases.     That   is,    the   decrease   of   clearance   increases   the 


INTERNAL  COMBUSTION  ENGINES 


363 


efficiency.  Of  course  this  increases  the  pressure  at  c,  hence  in- 
creasing the  compression  increases  the  efficiency.  The  amount 
of  compression  is  fixed  by  the  allowable  temperature  at  the  end 
of  compression  and  by  the  pressure  at  the  end  of  the  explosion. 
If  the  pressure  of  compression  is  excessive  the  temperature  may 
be  high  enough  to  cause  premature  explosion  while  if  high  with 
a  rich  gas  the  explosion  pressure  is  too  high.  In  practice  the 
pressure  at  the  end  of  compression  is  80  to  160  Ibs.  per  square 
inch  depending  on  the  kind  of  gas. 
As  before: 

Td-  Tb 


Td 


^3 

^2    =    ~ 


AW 


ATKINSON  CYCLE  AND  DIESEL  CYCLE 

Certain  other  cycles  have  been  proposed  for  gas  engines  and 
one  will  be  examined  for  the  purpose  of  application  of  theory. 

In  Fig.  173  the  cycle  proposed  by  Atkinson  is  shown.  In  this 
engine  pistons  were  so  connected  by  linkage  to  the  shaft  that  they 
made  strokes  of  varying  length. 
The  suction  stroke  ab  is  a  short 
stroke  followed  by  a  compression 
stroke  be.  The  explosion  cd  is 
followed  by  a  long  stroke  so  that 
the  expansion  reduces  the  pressure 
to  the  initial  value.  The  dis- 
charge stroke  ea,  which  is  long, 


FIG.  173. — Atkinson  cycle. 


brings  the  pistons  to  their  original  points. 
In  this  case  the  heat  added  is 


=  cv(Td  -  Te) 


The  heat  removed  is 


Q2  =  cp(Te  -  Tb) 
Work  =  Qi  -  Q2  =  cv[(Td  -  Tc)  -  k(Te 

,     Te  ~    Tb 


(7) 


364 


HEAT  ENGINEERING 


Here  there  is  no  simple  relation  between  the  temperatures  at 
the  corners  as  this  is  not  a  simple  cycle  of  poly  tropics. 

Rudolph  Diesel  in  the  last  decade  of  the  19th  century  pro- 
posed a  new  cycle  of  higher  efficiency.  He  recognized  the  fact 
that  to  obtain  high  efficiencies,  the  cycle  of  the  gas  engine  must 
approximate  the  Carnot  cycle;  that  the  range  of  temperature  on 
this  cycle  must  be  as  large  as  possible  and,  on  -account  of  the  loss 
of  availability  in  conduction,  the  engine  must  have  internal  com- 
bustion. He  proposed  a  cycle  of  isothermal  compression  from 
a  to  b,  Fig.  174,  followed  by  adiabatic  compression  be  to  such  a 
high  temperature  that  fuel  would  ignite  on  being  introduced  into 
the  cylinder.  The  fuel  was  to  be  introduced  at  such  a  rate  that 
its  burning  would  produce  just  enough  heat  to  make  the  expan- 
sion from  c  to  d  isothermal.  After  the  fuel  was  cut  off  the  ex- 
pansion da  became  adiabatic.  His  original  paper  in  the  Zeit- 


FIG.  174. — Original  Diesel  cycle. 


FIG.  175. — Final  Diesel  cycle. 


schrift  des  Vereins  Deutscher  Ingineure  was  translated  in  the 
Progressive  Age  in  Dec.,  1897  and  Jan.,  1898.  This  paper  gave 
the  results  of  his  work  for  a  number  of  years.  The  great  pres- 
sure developed  by  the  final  compression  and  the  slight  gain  of 
area  by  the  isothermal  ignition  led  him  to  abandon  the  upper  part 
of  the  figure,  while  the  desire  to  reduce  the  volume  led  to  cut- 
ting out  the  lower  end  of  the  figure.  Thus  modified  the  Diesel 
Cycle  took  the  form  shown  in  Fig.  175,  in  which  adiabatic  com- 
pression in  a  cylinder  of  small  clearance  brings  the  air  from  a  to 
b  at  which  the  temperature  is  high  enough  to  ignite  oil  which  is 
injected  into  the  cylinder.  If  the  burning  progresses  at  the 
proper  rate  the  pressure  will  be  kept  constant  until  cut  off  at  c. 
From  c  to  d  adiabatic  expansion  takes  place  followed  by  exhaust 
from  d  to  a  and  finally  to  e.  The  suction  stroke  from  e  to  a 


INTERNAL  COMBUSTION  ENGINES 


365 


charges  the  cylinder  with  air.     This  is  a  four-stroke  cycle.     The 
efficiency  is  given  by 

cP(Tc-Tb)-cv(Td-Ta) 


cp(Tc-Tb) 
Td-Ta 


k(Tc-Td) 
The  high  theoretical  efficiency  is  due  to  the  high  compression. 


(8) 


ACTUAL  ENGINES 


The  form  taken  by  an  actual  engine  is  shown  in  Fig.  176. 
This  represents  an  engine  of  the  Otto  form.  In  this  air  is  drawn 
into  the  cylinder  by  the  outward  motion  of  the  piston  A  through 
the  valve  B.  Gas  is  admitted  by  C  which  is  so  arranged  as  to 


FIG.  176. — Ordinary  gas  engine.     4-cycle. 

open  after  B  by  means  of  a  shoulder  formed  by  a  sleeve  attached 
to  the  valve  stem.  These  admission  valves  are  opened  by  the 
suction  of  the  piston  or  they  may  be  opened  positively  by  the 
valve  gearing.  They  are  closed  by  the  spring  D,  when  the  piston 
reaches  the  end  of  its  stroke  or  when  the  valve  mechanism  re- 
leases them.  At  the  end  of  the  expansion  stroke  the  exhaust 
valve  E  is  opened  by  the  lever  F  operated  by  a  cam.  The 
cam  is  on  a  cam  shaft,  operated  by  a  bevel  or  spiral  gear  from 
the  crank  shaft.  The  speed  of  this  cam  shaft  for  a  four-cycle 
engine  is  one-half  the  speed  of  the  engine  shaft.  The  governor 


366 


HEAT  ENGINEERING 


operates  to  throttle  the  mixture  or  to  prevent  gas  from  entering 
as  will  be  explained  later.  The  cylinder  K  contains  a  water 
jacket  L  for  the  purpose  of  keeping  the  temperature  of  the  cyl- 
inder wall  low  enough  for  lubrication  and  also  to  prevent  seizing. 
The  two-cycle,  Mietz  and  Weiss  oil  engine,  Fig.  177,  is  similar 
in  action  to  the  four-cycle  engine.  In  this  the  explosion  and 
expansion  of  the  charge  compresses  air  in  the  crank  case  K 
and  when  the  piston  A  overrides  the  port  B  the  burned  gases 
escape  to  the  exhaust  pipe  M,  while  at  the  next  moment  the 


FIG.  177. — Two-cycle  Mietz  and  Weiss  oil  engine. 

ports  C  are  uncovered,  admitting  the  air  compressed  in  the 
crank  case.  This  compressed  air  rushes  in  and  is  deflected  to  the 
head  by  the  projecting  finn  D  so  that  the  burned  gases  are 
blown  out  or  scavenged.  This  reduces  the  air  in  the  crank  case 
to  atmospheric  pressure.  After  the  piston  passes  C,  the  air  in 
the  crank  case  is  rarified,  as  no  air  can  enter,  and  after  passing 
B  the  air  in  the  cylinder  is  compressed.  When  the  piston 
travels  back  far  enough  to  uncover  port  E  air  is  drawn  into 


INTERNAL  COMBUSTION  ENGINES 


367 


the  crank  case  from  the  base  of  the  engine  by  the  partial  vacuum 
existing  there.  In  the  Mietz  and  Weiss  engine  kerosene  is 
sprayed  into  the  cylinder  near  the  end  of  the  stroke  by  the  pump 
G,  and  vaporized  by  the  hot  cylinder  head  H  and  finally  ignited 
by  the  high  temperature  from  compression  in  the  heated  ball 
F  at  the  end.  The  mixture  then  explodes  and  the  action  de- 
scribed above  is  repeated. 

In  small  engines  using  gasolene,  the  air  entering  the  crank 
case  is  drawn  through  a  carbureter  in  which  the  air  is  drawn 
through  gasolene  or  is  mixed  with  it.  This  charges  the  air  with 
fuel  and  the  mixture  is  ignited  at  the  end  of  compression  by  a 


Scale  120  Ib. .  l" 
Cylinder  Card 


Scale  10  Ib.  «=!' 
Crank  Case  Care 


FIG.  178. — Cards  from  cylinder  and  crank  case  of  a  two-cycle  engine. 

spark.  In  large  two-cycle  engines  the  fuel  gas  and  air  are 
compressed  in  piston  compressors  and  admitted  to  the  cylinder 
at  the  proper  time.  Fig.  178  illustrates  the  form  of  indicator 
cards  from  the  cylinder  and  crank  case  of  the  engine  above. 

GOVERNING 

Internal  combustion  engines  are  governed  in  two  principal 
ways:  (a)  by  the  hit  and  miss  system  and  (6)  by  the  throttled 
charge  system.  In  the  hit  and  miss  method  of  governing  the 
governor  acts  on  the  gas  or  fuel  valve  so  that  no  gas  is  admitted 
when  the  speed  exceeds  a  given  limit,  but  as  long  as  the  limit 
is  not  exceeded  the  engine  receives  its  full  charge.  In  throttle 
governing  the  charge  of  fuel  and  air  may  be  throttled  or  the 


368  HEAT  ENGINEERING 

fuel  may  be  throttled  alone.  In  the  first  of  these  less  fuel  mix- 
ture is  introduced  into  the  cylinder  although  it  is  of  the  proper 
mixture,  while  in  the  second  case  the  mixture  is  changed.  In 
each  of  these  the  explosion  pressure  will  be  reduced  and  the 
work  will  be  made  less.  Of  course  the  efficiency  in  throttle 
governing  is  reduced.  The  disadvantage  of  the  hit  and  miss 
system  is  the  fact  that  the  speed  may  fluctuate,  due  to  this 
method.  In  the  Diesel  engine  the  governor  fixes  the  point  at 
which  the  oil  supply  is  cut  off. 

IGNITION 

The  charge  in  most  modern  gas  engines  is  ignited  by  an  elec- 
tric spark  made  by  breaking  a  circuit  between  two  platinum  points 
or  else  causing  a  high-tension  spark  to  jump  a  gap  between  two 
points  of  platinum  or  tungsten.  The  method  of  compressing 
the  charge  into  hot  tube  until  the  mixture  comes  in  contact 
with  red  hot  iron  is  rarely  used  at  present  and  compression  to  a 
high  temperature  for  ignition  is  used  in  some  small  engines.  This 
latter  method  is  used  exclusively  on  the  Diesel  engine. 

The  rapidity  of  ignition  depends 
on  the  pressure  and  the  quality  of 
the  mixture.  Thus  with  higher  pres- 
sures the  ignition  is  more  rapid.  In 
many  cases  it  is  found  that  the  ex- 
plosion is  not  instantaneous  but  is 

continued  after  the  end  line  as  shown 
in  Fig.  179.  Such  action  is  called 
after  burning.  This  after  burning  is  found  to  take  place  in 
weak  mixtures  and  in  throttled  supply.  The  efficiency  of  the 
engine  is  found  to  increase  as  more  air  is  added  with  the  gas 
than  that  required  theoretically.  There  are  several  theories 
given  for  this,  a  satisfactory  one  being  that  as  the  heat  per 
cubic  foot  is  decreased  by  the  addition  of  air,  the  temperature 
increase  is  not  so  great  and  the  loss  from  the  cylinder  is  made 
less,  moreover  this  might  give  a  different  specific  heat  and 
thus,  a  lower  temperature.  The  after  burning  may  be  ex- 
plained by  the  fact  that  in  many  cases  the  high  temperatures 
present  prevent  chemical  combination  or  may  lead  to  dissociation. 
The  amount  of  dilution  by  excess  air  to  insure  the  presence  of 
sufficient  oxygen  is  a  matter  of  experiment.  The  excess  air  may 


INTERNAL  COMBUSTION  ENGINES 


369 


be  sufficient  to  make  the  excess  air  and  inert  nitrogen  amount 
to  about  5  times  the  gas  and  the  oxygen  required  to  burn  it, 
although  much  larger  variations  have  been  observed  in  explosive 
mixtures.  The  time  of  explosion  varies  with  the  amount  of 
excess  air  present  being  slow  with  large  quantities.  A  slight 
excess  gives  a  quick  explosion.  It  has  been  found  at  times  that 
the  best  efficiency  was  obtained  even  with  100  per  cent,  excess  air. 

HEAT  TRANSFER  TO  WALLS 

Experiments  have  been  made  to  determine  the  loss  of  heat  to 
the  cylinder  walls  in  a  similar  manner  to  that  used  for  the  steam 
engine.  Coker  found  that  there  was  a  cyclic  variation  of  7°  C. 
or  12.6°  F.  at  a  depth  of  0.015  in.  in  the  wall  of  a  cylinder  of 
an  engine  running  at  240  r.p.m.  In  the  Seventh  Report  of  the 
Committee  on  Gaseous  Explosion  of  the  British  Association, 
the  results  of  Dr.  Coker  and  Mr.  Scoble  are  shown.  Fig.  180  is 
taken  from  this  report. 


180  270  300  450  540 

Angular  Fgame  Of  Crank  in  Degrees 


630 


FIG.  180. — Cycle  in  gas  temperature  from  a  gas  engine  after  the  Gaseous 
Explosion  Committee  Report. 

This  curve  shows  the  temperature  of  the  working  medium  as 
determined  by  a  platinum  couple.  The  couple  was  placed  in  a 
tube  which  could  be  cut  off  from  the  cylinder  by  a  valve  until 
the  desired  time.  The  high  temperature  of  the  cycle  was  not 
measured  but  computed  because  at  this  temperature  the  couple 
would  melt  if  exposed  to  the  gas.  These  curves  show  the  great 
variation  in  temperatures  which  takes  place  in  the  gas  engine. 

24 


370  HEAT  ENGINEERING 

The  results  of  Coker  give  a  variation  from  620°  F.  to  3960°  F.  for 
one  test,  while  for  another  it  extended  from  390°  F.  to  3250°  F. 
The  losses  from  the  cylinder  walls  are  due  to  radiation  from 
the  high  temperature  gas  as  well  as  from  convection  currents. 
These  losses  are  affected  by  the  density  of  the  gas.  An  increased 
density  will  increase  the  heat  loss.  In  addition  to  this  fact 
that  the  loss  is  made  greater  by  increasing  the  density  through 
the  increase  of  compression,  the  danger  of  pre-ignition  through 
an  increase  of  the  temperature  from  this  cause  gives  a  limit  to 
the  possible  compression. 

FUELS 

The  fuels  used  in  internal  combustions  are  inflammable  gases 
and  oils  although  Diesel  proposed  in  his  patents  to  use  powdered 
solid  fuels.  The  gases  in  common  use  are  natural  gas,  illuminat- 
ing gas,  producer  gas  and  blast-furnace  gas. 

Natural  gas  is  a  product  of  nature  found  in  certain  localities 
in  various  parts  of  the  world,  usually  where  oils  are  found.  It 
is  rich  in  methane.  Its  heating  value  is  high.  T.  R.  Wey- 
mouth  in  the  Journal  of  the  A.S.M.E.,  May,  1912,  gives  the 
following  average  analysis  for  natural  gas  of  the  United  States: 

Methane,  CH4 87.00 

Ethane,  C2H6 6.50 

Ethylene,  C2H4 0.20 

Carbon  monoxide,  CO 0 . 20 

Hydrogen,  H2 Trace 

Nitrogen,  N2 5 . 50 

Carbon  dioxide,  CO2 0.50 

Helium,  He 0.10 

Oxygen,  O2 Trace 

100.00 

The  average  heating  value  of  the  gas  was  887.3  B.t.u.  per 
cubic  foot  at  29.82  in.  and  60°  F.  The  specific  gravity  com- 
pared with  air  was  0.6135. 

Illuminating  gas  is  only  used  for  small  installations  as  the 
cost  is  prohibitive.  Its  heating  value  is  about  700  B.t.u.  per 
cubic  foot  under  standard  conditions  of  760  mm.  and  0°  C.  This 
gas  may  be  the  product  of  distilling  bituminous  coal,  the  gas 
amounting  to  30  per  cent,  of  the  coal.  The  residue  is  coke  and 
unless  it  can  be  sold  this  method  of  utilizing  coal  is  not  ef- 
ficient. If,  however,  the  illuminating  gas  is  made  by  enriching 


INTERNAL  COMBUSTION  ENGINES  371 

producer  gas  by  adding  hydrocarbons  from  crude  oil  and  fixing 
the  mixture,  this  waste  of  coke  would  not  occur.     If  the  gas 
were  made  for  power  purposes  alone  this  enrichment  would  not 
be  made  as  producer  gas  is  satisfactory  for  use  in  gas  engines. 
An  analysis  of  illuminating  gas  is  given  below: 

Hydrogen,  H2 34.3 

Methane,  CH4 28.8 

Ethylene,  C2H4 9.5 

Heavy  hydrocarbons,  C2He 1.7 

Carbon  dioxide,  CO2 0.2 

Carbon  monoxide,  CO 10 . 4 

Oxygen,  O2 0.4 

Nitrogen,  N2 14.7 


100.0 

The  common  form  of  gas  used  in  power  installations  is  producer 
gas.  This  gas  is  made  from  any  form  of  coal  or  lignite  in  a 
producer  as  shown  in  Fig.  181  or  Fig.  182.  These  represent  two 
forms:  the  pressure  producer  and  the  suction  producer.  Fig.  181 
illustrates  one  form  of  pressure  producer  of  R.  D.  Wood  &  Co. 
In  this,  air  is  blown  into  a  bed  of  incandescent  fuel  A  by  means 
of  the  steam  jet  blower  B.  The  air  burns  the  carbon  to  CO2, 
but  on  passing  through  the  thick  bed  of  fuel  this  is  reduced  to 
CO  and  the  gas  rises  to  the  outlet  C.  The  heat  of  the  fire  serves 
to  drive  off  the  volatile  matter  from  the  coal. 

The  carbon  burning  to  CO  liberates  4450  B.t.u.  per  pound 
of  carbon  and  this  heat  would  be  lost  in  the  scrubber  if  not 
absorbed  in  some  manner.  Some  of  this  heat  is  used  to  make 
steam  in  the  top  of  the  producer  or  in  a  boiler  heated  by  the 
exhaust  gases  on  their  way  to  the  scrubber.  If  this  steam  is 
passed  into  the  ingoing  air,  the  dissociation  of  this  by  the  heat 
of  combustion  of  the  fuel  will  absorb  much  of  this  heat. 

The  steam  used  in  the  air  blast  cools  the  gas  as  it  is  disso- 
ciated into  hydrogen  and  oxygen  on  passing  over  the  hot  coals. 
This  dissociation  utilizes  part  of  the  heat  of  combustion  of 
carbon  to  CO  and  so  utilizes  some  of  that  which  would  be  lost 
in  the  scrubber.  The  steam  also  prevents  the  fire  from  clink- 
ering.  The  gas  now  contains  CO,  H2,O2  and  N2  as  well  as  some 
volatile  gases. 

Fresh  coal  is  discharged  from  the  hopper  D  into  the  chamber 
E  and  from  this  it  is  distributed  by  a  special  device  F  which 


372 


HEAT  ENGINEERING 


produces  in  a  uniform  bed.  The  openings  G  in  the  top  are  for 
the  introduction  of  slice  bars  to  break  up  the  fuel  bed.  The 
openings  H  on  the  side  are  for  observation  of  the  fuel  bed  and 
for  barring  the  bed  when  necessary.  The  producer  is  capped 
by  a  water-cooled  top  and  the  sides  are  lined  with  fire  brick.  The 
ashes  drop  into  a  water-sealed  base.  From  the  producer  the 
gas  passes  through  the  outlet  C  into  a  scrubber  which  consists 


FIG.  181. — Pressure  producer  of  the  R.  D.  Wood  Co. 

of  a  metal  cylinder  filled  with  coke  and  cooled  by  water.  The 
gas  enters  a  water  seal  at  the  bottom  of  the  scrubber  and  leaves 
at  the  top.  From  this  point  the  gas  may  be  taken  to  a  tar 
extractor  if  the  gas  contains  tar  to  any  extent.  This  is  practi- 
cally a  centrifugal  fan  over  which  the  gas  passes,  being  thrown  to 
the  outer  edge  and  thus  causing  the  heavier  tar  to  be  caught  by 
the  casing.  From  this  point  it  is  taken  to  the  gas  holder. 


INTERNAL  COMBUSTION  ENGINES 


373 


In  Fig.  182  the  Otto  Suction  Gas  Producer  is  shown.  In 
this  a  suction  is  produced  by  the  engine  drawing  in  gas.  This 
draws  air  from  the  atmosphere  at  A  over  the  hot  water  in  the 
top  of  the  producer  at  B  and  thence  through  the  pipe  R  to  the 
bottom  of  the  fuel  bed  C.  The  water  is  heated  by  the  gas  passing 
from  the  producer.  It  then  passes  through  the  down  pipe  D 
to  the  seal  box,  depositing  tar  or  dirt  at  the  bottom  and  passing 
up  through  the  scrubber  E;  it  finally  enters  the  receiver  F  and 
then  passes  to  the  engine. 


FIG.  182. — Suction  producer  of  the  Otto  Gas  Engine  Co. 


The  blower  G  is  used  in  starting  the  fire  in  the  producer.  The 
three-way  cock  H  is  used  to  direct  the  gases  and  smoke  to  the 
chimney  before  burnable  gas  is  produced  in  starting  a  fire.  The 
scrubber  E  is  filled  with  coke  over  which  water  trickles  and 
through  which  the  gas  passes.  The  spherical  charging  ball 
at  the  hopper  mouth  prevents  gas  discharging  or  air  entering. 

An  analysis  of  producer  gas  from  soft  coal  gave  the  following: 


374  HEAT  ENGINEERING 

Carbon  monoxide,   CO 20 . 9 

Hydrogen,  H2 15.6 

Carbon  dioxide,  CO2 9.2 

Oxygen,  O2 , 0.0 

Ethylene,  C2H4 0.4 

Methane,  CH4 1.9 

Nitrogen,  N2 52.0 

100.00 

This  gas  gave  156.1  B.t.u.  per  cubic  foot  under  standard 
conditions. 

Since  1900  blast-furnace  gas  has  been  used  successfully. 
Apparently  it  was  first  used  in  England  in  1895  on  a  12  X  30  — 
15-h.p.  gas  engine  and  in  the  same  year  on  an  8-h.p.  engine  in 
Belgium;  in  1898  a  150-h.p.  engine  was  used;  in  1899  a  600-h.p. 
engine  was  used  in  Belgium  and  exhibited  by  Cockerill  &  Co. 
at  the  Paris  Exposition  of  1900.  These  were  followed  by  larger 
engines.  In  1903  the  Lacka wanna  Steel  Company  installed  a 
number  of  2000-h.p.  engines  using  blast-furnace  gas.  In  most 
cases  the  gas  is  cleaned  by  centrifugal  washers  before  it  is  applied 
in  the  gas  engine.  This  gas  was  formerly  used  beneath  boilers 
for  steam  generation  but  gas  engines  utilize  the  heat  more 
effectively. 

An  analysis  of  blast-furnace  gas  is  given  herewith: 

CO 25.83 

CO2 9.37 

CH4 0.54 

H2.. 2.96 

N2 61.30 

100.00 

Heat  value  =  105  B.t.u.  per  cubic  foot. 

Crude  oil  is  usually  used  by  the  Diesel  engine.  This  oil 
varies  some  in  composition  and  heating  value.  The  analysis 
below  is  for  one  form : 

C 84.9 

H.. 13.7 

0 1.4 

100.0 

Its  heating  value  is  19,000  to  20,000  B.t.u.  per  pound. 

Gasolene  is  the  light  first  distillate  from  crude  petroleum  oil. 
It  is  easily  vaporized  and  in  gas  engines  it  is  usually  delivered  to 
the  air  supply  by  a  form  of  mixer  known  as  a  carbureter.  Its 


INTERNAL  COMBUSTION  ENGINES  375 

heating  value  is  about  20,500  B.t.u.  per  pound  when  the  steam 
formed  is  reduced  to  water.     It  contains  approximately: 

C , 83.5  percent. 

H 15.5  per  cent. 

N2,  S,  and  O2 1 . 0  per  cent. 

Kerosene  is  one  of  the  later  distillates  of  crude  mineral  oil. 
It  is  used  on  a  few  engines.  It  requires  a  special  form  of  car- 
bureter or  vaporizer,  requiring  a  high  temperature  to  properly 
volatilize  the  oil.  The  analysis  of  this  gives: 

Carbon 84^  per  cent. 

Hydrogen 14     per  cent. 

Nitrogen  and  oxygen Itf    per  cent. 

The  heating  value  will  be  about  19,900  B.t.u.  for  the  higher 
heating  value. 

COMBUSTION  OF  FUELS 

The  combustion  of  these  various  gases  is  accomplished  by  the 
mixing  of  air  with  the  fuel.  To  get  a  mixture  which  will  explode 
or  burn  rapidly  the  quantity  of  air  must  be  within  certain  limits, 
usually  the  amount  necessary  for  complete  combustion  plus  15 
per  cent,  will  give  good  results.  To  find  the  weight  or  volume  of 
air  to  burn  any  given  constituent,  reference  is  made  to  chemical 
formulae  and  Avogadro's  Law. 

Thus  to  burn  1  volume  of  CH4,  2  volumes  of  oxygen  are 
required — which  means  9.54  volumes  of  air.  From  this  burn- 
ing 1  volume  of  CO2  results  and  2  volumes  of  H2O.  The  water 
may  condense.  The  weights  of  these  per  pound  of  CH4  are: 
4  Ibs.  oxygen,  17.40  Ibs.  air,  2%  Ibs.  CO2  and  2%  Ibs.  of 
water.  These  statements  are  seen  from  the  following: 

CH4  +  202  =  C02  +  2H20 

By  Avogadro's  Law: 

1  volume  CH4  +  2  volumes  O2  =  1  volume  CO2  +  2  volumes 
H2O. 
By  chemical  equivalents: 

[12  +  4.032]  Ibs.  CH4  +  2[16  X  2]  Ibs.  02  =  [12  +  32]  Ibs.  CO2 
+  2[2.016  +  16]  Ibs.  H20. 
or 

1  Ib.  CH4  +  4  Ibs.  02  =  2%  Ibs.  C02  +  2%  Ibs.  H20. 


376  HEAT  ENGINEERING 

Air  contains  77  parts  by  weight  of  nitrogen  and  23  parts  oxygen 

while  by  volume  the  proportion  is  79  to  21.     Hence  2  volumes  of 

2 
oxygen  mean  TT-^T  or  9.54  volumes  of  air  and  4  Ibs.  of  oxygen 

O.Zl. 
4 
mean  ^-^  =  17.40  Ibs.  of  air. 

U.^o 

The  heat  of  combustion  of  CH4  is  21,566  B.t.u.  per  pound  if 
the  water  produced  remains  as  steam  or  24,019  B.t.u.  per  pound 
if  the  steam  is  condensed  to  water.  The  former  is  spoken  of 
as  the  lower  heating  value,  the  latter  as  the  higher. 

In  gas-engine  work  in  England,  Germany  and  America,  the 
lower  value  is  taken  in  determining  the  efficiencies,  while  in 
France  the  higher  value  is  taken.  It  is  fairer  to  use  the  higher 
heating  value  and  charge  the  engine  with  the  heat  which  is  pres- 
ent with  the  steam  in  the  exhaust. 

To  find  the  amount  of  heat  per  cubic  foot  of  gas  the  numbers 
above  would  be  divided  by  the  volume  of  1  Ib.  of  the  gas 
under  certain  definite  conditions.  .  The  conditions  assumed  in 
many  cases  are  14.7  Ibs.  per  square  inch  pressure  and  32°  F. 
although  for  natural  gas  the  conditions  are  often  a  pressure  of 
28.7  in.  of  mercury  and  60°  F.  For  the  former  case  the  volume 
of  1  Ib.  of  gas  is  given  by: 


(9) 


MBT       MET      __1 1  X  mol.  wt.  > 

p  P  mol.  wt.  14.7  X  144 

359 


mol.  wt. 


The  specific  weight  is  given  by: 


1       mol.  wt. 
m  =  v  = 


For  CH4:  v  =  =  22.4  cu.  ft. 


This  gives  the  values  of  1072  B.t.u.  per  cubic  foot  for  the  greater 
heating  value  and  963  B.t.u.  for  the  lower  value  for  CH4. 

Using  the  methods  above  for  various  gases  and  elements  the 
following  table  is  computed: 

GAS  COMPOSITION 

To  study  the  action  of  the  gas  engine  theoretically  the  actual 
gas  supplied  and  the  amount  of  air  or  the  products  of  combustion 


INTERNAL  COMBUSTION  ENGIh 

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378 


HEAT  ENGINEERING 


must  be  known.  From  these  the  various  temperatures  and  forms 
of  lines  may  be  determined.  From  the  chemical  analysis  of  the 
fuel  gas  and  the  exhaust  gas,  the  amount  of  air  may  be  found,  as 
will  be  shown  below. 

Suppose  then  that  the  producer  gas  given  on  p.  374  is  to 
be  burned  in  a  gas  engine  with  25  per  cent,  dilution  and  it  is  re- 
quired to  know  the  amount  of  air  to  be  admitted,  the  volume  of 
the  mixture,  the  heat  per  cubic  foot  of  mixture  and  the  products 
of  combustion. 


Per  cent. 
Vol.  gas 

Volume 
air 

Heat  of 
combus- 
tion 

Products  of  combustion 

C02 

02 

N2 

H2O 

SO2 

CO  
H2  
CO2  ... 
O2  

20.9 
15.6 
9.2 
0.0 
0.4 
1.9 
52.0 

49.9 

37.2 

7,060 
5,130 

20.9 
9.2 

39.3  . 
29.5 

15.6   . 

0.0 
5.7 
18.1 

0.8   . 
3.8   . 

C2H4... 
CH4.... 

N2  

Excess 

669 
2,037 

0.8 
1.9 

4.52 
14.30 
52  00 

5.82 

21.30. 

100.0 
Air.... 

110.9 

27.7 

14,896 

138.6 

14,896 

32.8 

5.82 

160.92 

20.2 

This  table  has  been  prepared  from  table  on  p.  377  for  100  Ibs. 
of  gas  in  the  following  manner: 

Air  for  CO  =  20.9  X  2.38  =  49.9 
High  Heat  for  CO  =  2.09  X  337  =  7060 
CO2  from  CO  =  20.9  X  1.0  =  20.9 
N2  from  CO  =  20.9  X  1.88  =  39.3 

14896 
B.t.u.  per  cubic  foot  mixture  = 

1  S8  6 
Air  per  cu.  ft.  -        =  1.386 


Vol.  of  mixture  =  2.38  cu.  ft. 
Products  of  combustion  per  cubic  foot: 

CO2  =  0.328  cu.  ft. 
O2  =  0.058  cu.  ft. 
N2  =1.609  cu.  ft. 

The  values  of  B,  cp  and  cv  for  a  mixture  are  important  to  con- 


INTERNAL  COMBUSTION  ENGINES  379 

sider.  Since  the  molecular  weights  of  the  various  constituents 
are  different  from  each  other  and  since  the  mixture  contains  CO2 
and  water  vapor  the  expressions  for  specific  heat  must  be  com- 
puted. The  a's  of  the  expression 

cv  =  a  +  bt 

are  different  for  these  various  substances  and  the  same  is  true 
for  a''s  and  b's.  If  the  values  of  a,  b  or  a'  for  1  Ib.  of  a  gas  be  di- 
vided by  the  volume  of  1  Ib.  under  standard  conditions  the 
value  of  these  quantities  for  the  heat  to  be  added  to  1  cu.  ft.  of  a 
substance  to  change  its  temperature  1  deg.  will  be  found.  This 
might  be  called  the  specific  heat  of  I  cu.  ft.  The  reason  for  using 
this  quantity  is  the  fact  that  the  gas  analysis  given  is  usually  by 
volume  and  the  necessity  of  reducing  this  to  percentage  by 
weight  is  eliminated.  This  is  done  with  the  heat  of  combustion 
to  reduce  that  to  the  heat  per  cubic  foot. 

If  Vp  is  the  percentage  volume  or  the  relative  volume  of  the 
various  gases  in  a  mixture  the  following  equations  hold  for  the 
mixture: 

vp  =  partial  volume  of  any  constituent. 

Mol.  Wt.mix  -  Zfo  X^ut.).  .  (11) 

a-  =  ^-*  ™ 

bmix _  ^m     .  (13) 

0^  .  z^_x^).  (14) 

2i>  \J-^:/ 

•    H  2(»pXff). 

ix  =       ~2v —  ^     ' 

Cpmix  =  amix  -f-  bmixT.  (16) 

Cvmix  =  a'mix  +  bmixT.  (17) 

B  ^^~ 

Mol.Wt.mix 

359  E_ 

~  Mol.wt.  ~  4.3' 

(20) 


380 


HEAT  ENGINEERING 


PROBLEM 

Suppose  the  natural  gas  given  on  p.  370  was  used  in  a  gas  engine  and 
it  is  required  to  find  the  amount  of  air  used  per  cubic  foot  of  gas  and  the 
true  products  of  combustion  if  the  exhaust  gas  analysis  by  volume  is : 

CO2  =  8.8  per  cent. 
N2  =  85.7  per  cent. 
O2  =  5.5  per  cent. 

These  analyses  show  no  water  vapor  from  the  burning  of  the  gas  nor  from 
that  taken  in  with  the  air  and  gas.  Suppose  the  gas  is  at  80°  and  saturated 
while  the  air  is  taken  at  70°  and  is  1A  saturated. 

The  pressure  of  the  water  vapor  in  the  gas  is  0.5056  Ib.  per  square  inch, 
while  that  in  the  air  is  0.1814  Ib.  per  square  inch.  The  barometer  is  14.7 
Ibs.  per  square  inch.  Now  from  Dalton's  Law  the  moisture  percentage 
by  volume  is  equal  to  the  percentage  of  the  dry  air  pressure  which  gives 
the  partial  pressure.  Hence 

Per  cent,  volume  of  gas  entering  equal  to  moisture  = 

0.5056 


14.7  -  0.5056 

Per  cent,  volume  of  air  entering  equal  to  moisture  = 

0.1814 


3.56  per  cent. 


On  account  of  pressure  and  temperature  differences  of  gas  and  air  the 
theoretic  amount  of  air  per  cubic  foot  must  be  multiplied  by 

14.7  -  0.5056      460  +  70 


14.7  -  0.1814      460  +  80 


_ 

~ 


to  give  the  actual  cubic  feet  of  air  per  cubic  foot  of  gas. 

The  true  constituents  of  the  gas  are  those  given  on  p.  370  multiplied  by 
0.964  to  which  3.56  per  cent,  moisture  is  added.  These  quantities  are  then 
multiplied  by  the  quantities  of  the  table  on  p.  377  to  give  the  various 
volumes  of  the  products  of  combustion  and  the  volumes  of  the  air  required. 
The  results  are  expressed  as  a  percentage.  If  the  total  air  required  is 
multiplied  by  1.25  per  cent,  the  moisture  from  the  air  is  found. 

The  computation  is  now  made. 


Per  cent, 
volume 

C02 

N2 

H20 

Air 

Heat 

CH4     87.00X0.964 
C2H4 

83.90 
0  19 

83.90 
0.38 

632.0 
2.15 

167.80 
0.38 

800.00 
2.72 

90,100 
318 

C2H6 

6  23 

12  46 

82.50 

18.69 

104.25 

11,420 

CO  

0.19 

0.19 

0.35 

0.45 

64 

N2  +  He 

5  41 

5  41 

CO2 

0  48 

0  48 

H2O  . 

3  60 

3.60 

100.00 

97.41 

722.41 

190.47 

907.42 

101,902 

INTERNAL  COMBUSTION  ENGINES  381 

Moisture  from  air  =  907.42  X  0.012  =  10.90 
Total  moisture  =  201.37 

The  volume  of  the  mixture  before  burning  is 

100  +  907.42  +  10.9  =  1018.32 
while  after  burning  the  volume  is 

97.41  +  722.41  +  201.37  =  1021.19 

There  is  a  slight  increase  in  volume. 

Products  of  combustion  in  the  original  analysis  show  that  there  is  some  air 
present.     The  amount  of  oxygen  is  5.5  per  cent.     This  is  associated  with 

5.5 

~-   X  0.79  =  20.6  per  cent,  of  nitrogen. 


The  nitrogen  in  the  analysis  from  combustion  must  be  the  difference 
between  85.7  per  cent,  and  20.6  per  cent. 

Nitrogen  from  combustion  =  85.7  —  20.6  =  65.1  per  cent.  The  nitro- 
gen in  the  products  of  combustion  give  a  total  of  722.14  parts  of  which  5.41 
have  been  brought  in  by  the  fuel  gas,  or 

5  41 
72^14  X  65.1  =  0.49 

is  the  amount  of  nitrogen  chargeable  to  gas.     The  nitrogen  from  the  air 
required  to  burn  the  gas  is  given  by  : 

65.1  —  0.49  =  64.61  per  cent.  =  volume  of  nitrogen 

from  air  to  burn  gases. 

85  7 
The  total  nitrogen  is  ~^  times  the  nitrogen  from  combustion,  or. 

85.7 


The  free  a'    '     2°'6 


65  l  X  722.41  =  952 


64.61    • 

Total  air  is  907.4  +  289.6  =  1197. 

Moisture  with  free  air  =  0.012  X  289.6  =  3.48. 

The  products  of  combustion  are : 

CO2 97.41  7.42 

N2 952.00  72.40 

5  5 
O2          ^-5X97.41 61.10  4.62 


H20              ;•":      204.85  15.56 

I       o .  48  J  

1315.36  100.00 
As  a  check,  the  total  volume  of  the  mixture  before  burning  is : 

Volume  gas 100 . 0 

Volume  air 1 197 .  Q 

Volume  moisture 14.4 

Total..                                                                 .  1311.4 


382  HEAT  ENGINEERING 

The  theoretical  air  per  cubic  foot  is 

^~  =  9.07  cu.  ft. 

The  theoretical  air  at  atmospheric  conditions 

9.07  X  0.96  =  8.71  cu.  ft. 
The  actual  amount  is 

1197  X  0.96 

1QQ =  11.49  cu.  ft. 

The  heat  per  cubic  foot  of  mixture  entering  is 
V  '  ^££  -  77.6  B.t.u.     |      "  -I?         • 

The  specific  heat  of  the  gas  mixture  entering  is  given  by  the  following ; 

Volumes 

Air... 1197.00 

CH4 

C2H4.. 


CO 

N2  +  He. 
CO2.. 


83.90 

0.19 

6.23 

0.19 

5.41 

1292  .  92 

0.48 

0.48 

3.60 

14.40 

18.00 

1311.40 

H20 | 


For  the  total  volume  of  1311.20  units  the  following  is  true: 

Va  Vb 

1292.92  X  0.018  =  23.3         1292.92  X  0.185  =  239.0 
0.48X0.0202=  0.0098         0.48X0.807=   0.3875 
18.00  X  0.0188  =  0.339         18.00X0.668=  12.0 

23.6488  251.3875 

Va' 

1292.92  X  0.0125  =  16.2 
0.48  X  0.0147  =    0.007 
18.00  X  0.0132  =    0.238 
16.445 

VCP  =  23.6488  +  251.39  X  10~5  T 
For  one  unit  volume 

Cp  =  0.0180  +  0.191  X  10~5  T 
VCV  =  16.445  +  251.29  X  10~6  T 

Cv  =  0.0125  +  0.191  X  10~6  T 
For  T  =  1000°  F.  these  become 


VCV  =  18.9554 

26.1588 


C,  ~  18.9554 


=  1.38 


INTERNAL  COMBUSTION  ENGINES 

The  B  for  this  mixture  of  gas  and  air  is  found  by  (18) 


383 


Air                    

J 
Volume 

1197.00     X 

Molecular        Relative     Per  cent, 
weight             weight         weight 

28.8          =     34,420           94.4 
16.032      =        1,344              3.7 
28.032      =              5           
30.048      =           189              0.5 
28.0                          5         
28.02        =           151              0.4 
44.0                        21              0.1 
18.016      =          324              0.9 

83.90     X 
0.19     X 

C2H4 

C2H6  
CO  

6.23     X 
0.19     X 
5.41      X 

N-  +  He 

CO2  
H20  

0  .  48     X 

18.00     X 

36,464          100.00 
_               36464         _ 

IVlOl.   Wt.OT    ~~    -I  o-i  -I 

1544 

40  ~ 
=  55.7 

Va 
77.02  X  0.018     =  1.387 
7.42  X  0.0202  =  0.1496 
15.56  X  0.0188  =  0.2925 

T 
T 

"m  ~  27.78 
For  the  exhaust  gases  : 

N2  72.40 
O2                         4.62       77.02 

C02  
H20...... 

77.02  X 
7.42  X 
15.56  X 

VCP  =    1.829 
Cp  =  0.0182 

..     7.42         7.42 

....    15.56       15.56 

1.8291 

Va' 
77.02  X  0.0125  =  0.965 
7.42  X  0.0147  =  0.109 
15.56  X  0.0132  =  0.205 

100.00 
Vb 
0.185  =  14.25 
0.807  =    5.99 
0.668  =  10.35 

30.59 

+  30.59  X  10~5  T 
+  0.306  X  10~5  T 

1.279 

VCV  =    1.279  +  30.59  X  10~5 
Cv  *=  0.0128  +  0.306  X  10~5 

ForT  =  2000°F.;FCP  =  2.441;  VCV  =  1.891;  k  =  —  - 

j.  .oy  J. 


1.29 


By  the  method  used  for  the  mixture,  the  B  for  the  ignited  gas  is  B  =  55.7. 
Although  this  value  of  B  is  the  same  as  that  for  the  unburned  mixture,  the 
composition  is  so  different  that  the  expressions  for  specific  heats  are  not  the 
same. 

If  mixtures  on  the  compression  and  expansion  are  those  assumed,  the 
values  of  the  various  quantities  are  as  follows : 

For  compression : 

Cpm  =  0.0180  +  0.191  X  10~5  T 
Cvm  =  0.0125  +  0.191  X  10~5  T 
For  1000°  F.,  a  mean  temperature: 
Cp  =  0.0199 
Cv  =  0.0144 


384 


HEAT  ENGINEERING 


0.0199 


=  1.38 


0.0144 
Bm  =  55.7 

H  =  77.6  B.t.u.  per  cu.  ft. 
v  =  12.95 

For  the  exhaust  gases : 

Cpm  =  0.0181  +  0.295  X  10~5  T 
Cvm  =  0.0128  +  0.295  X  10~5  T 

For  T  =  2000°  F.: 


Cpm  =  0.0239 
Cvm  =  0.0186 
0.0239 
CLOI86 
Bm  =  55.7 
v  =  12.95 


k  = 


1.283 


TEMPERATURES  AT  CORNERS  OF  CARD 

The  various  temperatures  at  the  corners  of  the  indicator  card 
of  the  gas  engine  cycle  are  now  computed  theoretically.  In 
this  computation  the  variation  of  specific  heat  will  be  disregarded 
at  first  after  which  the  effect  of  this  variation  will  be  investigated. 
In  a  gas  engine  the  exhaust  gas  which  remains  in  the  cylinder 
warms  the  incoming  gas  and  changes  its  temperature.  For 

this  reason  the  amount  of  gas 
actually  used  by  an  engine  cannot 
be  told  by  the  change  in  volume. 
Moreover  the  heat  removed  by 
the  cylinder  jacket  during  the 
different  events  affects  the  results. 
Consider  the  card  of  Fig.  183.  On 
the  suction  stroke  5-1  a  charge  of 
gas  and  air  is  drawn  in,  mixing 
with  the  burned  gases  of  volume 
F5  in  the  clearance  space.  The 

temperature  of  the  gases  in  the  clearance  space  T5  will  be  equal 
to  the  temperature  resulting  from  the  expulsion  of  the  exhaust 
gases  from  point  4.  The  gas  which  remains  in  the  cylinder  at  4 
acts  on  the  gas  driven  out  to  force  it  into  the  atmosphere  and 
hence  it  may  be  considered  to  expand  adiabatically  in  driv- 
ing out  the  exhaust.  The  gas  in  contact  with  the  piston  from 
1  to  5  may  be  considered  to  be  at  a  constant  temperature  equal 


FIG.  183. — Theoretic  form  of  in- 
dicator card. 


INTERNAL  COMBUSTION  ENGINES  385 

to  that  due  to  adiabatic  expansion  from  4  to  1  if  no  loss  or  gain 
of  heat  from  the  cylinder  walls  be  considered.     Thus  : 


Hence  T,  =  T,  "  (23) 

The  mixture  of  this  burned  gas  of  weight 

M 
M0 


and  the  fresh  charge  will  result  in  a  volume  of  gas  V\  at  a  pressure 
Pi  and  at  a  temperature  TV  The  weight  of  this  mixture  of 
fresh  air  and  gas  and  the  burned  gas  will  weigh  M  2  pounds. 


BTl 

Although  the  values  of  B  are  not  quite  the  same  in  these  two 
formulae  they  may  be  considered  the  same  in  this  work. 

If  this  gas  enters  the  cylinder  from  5  to  1  the  work  done  by 
the  entering  gas  on  the  piston,  pi(Vi  —  F5),  will  just  equal  the 
work  done  by  the  atmosphere  in  forcing  the  air  into  the  cylinder 
so  that  this  need  not  be  considered  in  equating  the  energy  at  5 
plus  the  energy  in  the  air  entering  to  the  energy  at  1. 


(M,  -  M0)  JcvTa  =  (24) 

PS  =  Pi. 

Tr  T 

* 


0.4  "  \BTl       BT,  I 

pi(Vi-  Vi)  _  Pi  (Vi  _  Vj\ 
OATa          0.4  V77!      Tj 

Now  the  clearance  V%  or  V&  is  given  as  I  times  the  displace- 
ment of  the  piston. 

F5  =  Z  [7i  -  V6]  (26) 

v*  =       Vl  (27) 


„ 

Hence 

25 


386 


HEAT  ENGINEERING 


•*-  1    — 


T6  +    lTa 

Substituting  for  T&  its  value  from  (23),  T\  reduces  to 


(28) 


Now 


(29) 


(30) 


(31) 


(32) 


Now  the  heat  added  from  2  to  3  is  called  VH  and  is  equal  to 
VCV(T3  -  Tz)  hence 

VT-f  ff 

*T!  +  £  (33) 


vcv 


Hence  T4  =  Ti  +  ^ 

These  equations  reduce  equation  (29)  to 


(34) 


\T1+    H'\ 

a 

k-1 

h    +lTa 

L  •  h  c,a\ 

i    H 

[  a  +  c.rj 

(35) 


In  this  equation  the  only  unknown  is  TI  but  the  equation  is 
implicit  so  that  its  solution  is  best  made  by  trial.  After  TI 
is  found  the  other  temperatures  may  be  found  in  succession. 
Thus  suppose  in  an  engine  with  25  per  cent,  clearance,  the  out- 
side gas  is  at  a  temperature  of  70°  F.  and  the  heat  per  cubic 
foot  is  80  B.t.u.  while  the  value  of  Cv  per  cubic  foot  is  0.013. 


1.25 


°-4 


1.9 


INTERNAL  COMBUSTION  ENGINES  387 

-P_  80        if  1.9 


JU«f 

[1.9 

80 

r  0.013  X  TiJ 

[TI  + 

80        1 

1.9 

113 
*    +(0.25X530) 

0.013  X  1.9J 

10+         8° 

^0.013  X  T, 

T,  = 


Since  the  last  term  in  the  denominator  is  small  the  value  of 
TI  is  practically  equal  to  (1  +  l)Ta  although  on  account  of 
another  term  being  additive,  the  value  will  be  slightly  smaller. 
(1  +l)Ta  =  1.25  X  530  =  662.  Hence  try  TI  =  625. 

1.25  X  530  X  3855  X  0.593 

3855  X  0.593  +  132 
If  627  is  tried 

1.25  X  530  X  3857  X  0.618 

3857  X  0.618  +  132 
Use  626.        T2  =  626  X  1.9  =  1190 

Qft 

T3  =  1190  +  =  7340 


These  are  all  higher  than  the  values  found  in  practice  due  to 
the  fact  that  the  combustion  on  the  line  2-3  is  not  always 
complete  at  the  point  3,  that  the  jacket  removes  heat  from  the 
cylinder  and  also  because  the  specific  heat  varies  with  the  tem- 
perature increasing  perceptibly  at  the  higher  temperatures. 
If  the  heat  of  combustion  from  2-3  be  reduced  to  75  per  cent. 
of  its  value  the  temperature  T3  will  be  materially  decreased. 

ADIABATICS 

The  compression  and  expansion  lines  of  the  card  have  been 
assumed  to  be  adiabatics  of  the  form 

pyi.4  =  const.  (26) 

This  assumes  that  the  gases  are  perfect  gases  and  that  cp  and 
Cv  are  constant.     Now  the  true  forms  for  the  specific  heat  are 

cp  =  o  +  bT  (16) 

and  cv  =  a'  +  bT  (17) 


388  HEAT  ENGINEERING 

and  the  equation  of  the  adiabatic  is 

0  =  cvdt  +  Apdv  =  (a  +  bT)dt  +  Apdv 

f  +  Mt=-ABd^  (36) 

In  the  above  expressions  the  quantities  refer  to  1  Ib.  al- 
though as  will  be  seen  later  it  might  be  simpler  to  have  them 
refer  to  1  cu.  ft. 

The  integration  of  this  between  limits  finally  gives 

a'  log,  ^  +  b(T2  -  rj  =  -  AB  loge  ^  (37) 

From  this  T2  may  be  found  if  7\  and     -  are  known. 


,        2       ,         2  2 

log  —       log  T-  -  log  TF 


(38) 


log  ^  log  ^ 

since  &  =  I?  p 

Pi       Ti  72 
In  a  form  involving  T2  and  T7!  only,  this  may  be  written 

AB  log,  p  +  a'  loge  p  +  6(  T2  -  !Ti) 
n  =  -  ^  (39) 


Of  course  since  —  must  be  known  to  find  T2  and  7\  this  form 
v\ 

is  not  necessary,  except  in  cases  in  which  it  is  desired  to  know 

the  value  of   n  over  a  given  range  of  temperatures.     This  is  a 

/•> 

closer  value  of  n  for  the  range  from  I  '2  to  TI  than  the  value  —  • 

cv 

Equation  (37)  may  be  written 

a!  log,  T+bT  +  AB  log,  V  =  const.  (40) 

a'  log,  p  +  a'  loge  V+  AB  \oge  V  +  bT  =  const. 
a'  loge  p  +  a  loge  F+  6T7  =  const. 

bpp 

pa/ya^r  =  pa/yo€  B   =  const.  (41) 

These  equations  considering  the  variation  of  cv  with  tempera- 
ture will  cut  down  the  pressure  of  compression  and  its  tempera- 
ture and  thus  affect  the  expression  for  efficiency  of  the  air  cycle. 


INTERNAL  COMBUSTION  ENGINES  389 

EFFICIENCY  CHANGE 

The  change  in  efficiency  due  to  this  change  in  c0  may  be  com- 
puted as  follows: 


cv 


AB  n          w 
—,  (1  -  ,3)  log. 


~  =  ~cT- 1  ~c~          L  loge  \T~^U  J  ^44^ 


This  expression  would  give  the  fractional  variation  of  effi- 
ciency —  -,  due  to  the  fractional  change  in  the  value  of  cv,  (  —  -)  • 

^?3  \     Cy  I 

TEMPERATURE  AFTER  EXPLOSION 

To  find  the  temperature  after  burning,  on  the  assumption  that 
no  heat  is  taken  away;  the  intrinsic  energy  after  burning  is 
made  equal  to  the  heat  per  cubic  foot  plus  the  intrinsic  energy 
before  burning.  If  there  is  an  assumed  loss  of  heat  to  the  jackets 
of  20  per  cent,  of  the  heat  of  combustion  this  same  method  is 
used  with  80  per  cent,  of  the  heating  value  of  the  fuel. 

U2  =  Ui  +  H  X  (100  -  per  cent,  loss  to  jackets)       (45) 
The  intrinsic  energy  at  any  point  is  given  by 


CT  bT2 

U  =  I  cvdt  =  a'T+  —-  (46) 

Hence 

a'T2  +  ^  +  #(1  -  loss  to  jackets)  =  a'iTs  +  |  7Y      (41) 

The  values  of  a'T*  -f  ^7Y  and  of  a\T3  +  ^TV  are  sometimes 

£  A 

plotted  as  shown  in  Fig.  184  so  that  for  a  given  temperature  T2 
the  heat  contained  in  the  gas  is  read  from  the  figure.  After  add- 
ing H  or  a,  portion  of  H  to  this,  the  curve  for  the  burned  mixture 


390 


ENGINEERING 


will  give  TS  corresponding  to  the  heat  contained  in  each  pound, 
or  cubic  foot,  the  intrinsic  energy. 

To  find  the  temperature  at  4  equation  (37)  is  used. 

If  now  the  mixture  at  point  1  is  assumed  to  be  at  626°  abso- 
lute, the  temperature  at  the  various  points  may  be  found  by 
the  methods  given  above.  Assume  the  gases  are  those  mentioned 
on  p.  380  and  that  the  clearance  is  25  per  cent.  Of  course  7\ 
would  be  worked  out  by  equation  (35)  if  not  known.  Using 


Intrinsic  Ettergy  per  Cu.Ft.inB.t.u. 

g  fe  §  §  8 

/ 

/ 

/ 

/ 

/ 

s 

/* 

^ 

s^ 

0  500°  F      lOOO'F       1500°F     2000°F      2500°F     3000°F      3500F    4000°F 

Abs.        Abs.         Abs..       Abs.         Abs.        A.bs.        Abs.       Abs. 
Degrees  Absolute 

FIG.  184.  —  Plotting   of   intrinsic    energy   for   variable    specific    heat. 

data  for  the  mixture  on  p.  383  in  equation  (37),  the  following 
results  : 

12.95  X  0.0125  X  2.3  log         +  12.95  X  0.191  X  10~5  [T2  -  626] 

X  55.7  X  2.3  log  5 


log 


X  10~5  Tz  =  2.797  +  0.0417  +  0.309  =  3.148 
Assume  log  T2  =  3.148 
T2  =  1400 


This  is  of  course  too  large  due  to  the  second  term. 

Assume   T2  =  1200 
3.080  +  0.080  =  3.160 

slightly  too  large  Try  T2  =  1150 

3.061  +  0.077  =  3.148 

This  is  correct. 


INTERNAL  COMBUSTION  ENGINES  391 

The  value  of  n  by  equation  (38)  is 

1160  . 

_  lQg626"+1°g5  _  0.265  +  0.699  _ 

71    —  1  r  —  f^  r*f\f\  —    -I  •«5O 

log  5  0.699 

C 

This  is  the  same  value  found  for  the  ratio  of  -—•  which  was  given 

cw 

as  1.38.     That  the  value  of  1150  may  be  seen  to  be  correct  the 
following  computation  is  made  by  equation  (4). 

/v.\   n-  I  /5\  0.38 

n-fip)      -026m    =1150 

\V-2/  \1/ 

To  find  Ts  the  correct  method  is  to  equate  the  intrinsic  energy 
at  3  to  that  at  2  plus  the  heat  of  combustion.  Since  the  composi- 
tion of  gas  after  burning  is  not  the  same  as  that  before,  the 
specific  heat  is  not  constant  during  burning  nor  are  a'  and  6 
constant  during  this  change.  Hence  the  above  method  is  the 
only  one  which  may  be  used  to  find  TV 

a'mT*  +  |TV  +  H  =  a'tft  +  ^7V 

1 2  Q5 
12.95  X  0.0125  X  1150  +  — ~  X  0.191  X  10~5  (1150)2  +  12.95 

Zi 

10  OF; 

X  77.6  X  0.80  =12.95  X  0.0128  XT  3  +  — ^—  X0.306X10"5  TV 

TV  +  8360r3  =  50,700,000 
T3  +  4180  =  8260 

T3  =  4080°  F. 

T4  is  found  by  equation  (37)  used  for  the  determination  of  TV 
12.95  X  0.0128  X  2.3  X  log  ^§2  +  12.g5  x  0.295  X  10~5 

X  [4080  -  Tt]  =  ^  X  55.7  X  2.3  log  5 
log  T4  -h  10.04  X  10~5  T4  =  3.610  +  0.408  -  0.301  =  3.717 

For  log  T74  =  3.71 
T74  =  5150 
This,  of  course,  is  too  large. 

Try  T4  =  3000 
3.478  +  0.302  =  3.780 


392 


HEAT  ENGINEERING 


Too  large. 
Too  small. 

Too  large, 
which  gives 

n 


log 


Try  T4  =  2500 
3.399  +  0.252  =  3.651 

Try  T,  =  2750 
3.440  +  0.280  =  3.720 

Try  T,  =  2740 
3.438  +  0.275  =  3.713 

4080 

0.172  +  0.699 


2745 


log5 


log  5 


0.699 


1.23 


In  the  calculation  for  the  value  of  n  on  the  two  lines,  it  has 
been  found  that  on  the  compression  line  n  =  1.38  while  on  the 
expansion  line  n  =  1.23.  From  actual  tests  as  mentioned  on  page 
214,  n  =  1.288  on  compression  and  1.352  on  expansion.  The 
value  of  n  is  found  by  plotting  the  logarithms  of  the  volumes 
and  pressures  of  various  points  and  drawing  a  straight  line 
the  slope  of  which  will  give  the  value  of  n.  These  temperatures, 
626,  1150,  4080  and  2745,  are  not  much  higher  than  those  found 
in  practice. 

TEMPERATURE  ENTROPY  DIAGRAM 

The  temperature  entropy  diagram  for  a  gas-engine  cycle  may 
be  easily  constructed  from  the  indicator  diagram  by  the  method 


3.00 


.S2.00 


1.0 


1.0 


2.0  3.0  4.0 

Volume  ID  Inches 


5.0 


6.0 


FIG.  185. — Card  prepared  for  T-S  analysis. 

shown  below.  The  mean  indicator  card,  Fig.  185,  is  constructed 
from  a  number  of  cards  in  the  same  manner  as  that  used  in  the 
Hirn's  or  Temperature  Entropy  Analysis,  and  then  the  lines  of 
absolute  zero  of  volume  and  of  pressure  are  laid  off  after  meas- 


INTERNAL  COMBUSTION  ENGINES 


393 


uring  the  clearance  and  calibrating  the  indicator  spring.  A 
series  of  points  is  then  marked  and  the  distances  from  the  axes 
are  measured  in  inches.  These  are  tabulated  as  shown: 


Points 

Pres- 
sure in 
inches 

Vol- 
ume in 
inches 

Px 

Pi 

V, 

vl 

log-^ 

Pi 

loglr 

*'osff 

S*  -  Si 

Tx 
Ti 

1 

0.15 

6.00 

1.00 

1.000 

0.000 

-0.000 

-0.000 

0.000 

.000 

2 

0.178 

5.25 

1.18 

0.873 

0.072 

-0.060 

-0.084 

-0.012 

.030 

3 

0.224 

4.43 

1.49 

0.737 

0.173 

-0.132 

-0.185 

-0.012 

.098 

4 

0.282 

3.75 

1.88 

0.625 

0.274 

-0.204 

-0.286 

-0.012 

.175 

5 

0.350 

3.18 

2.33 

0.530 

0.367 

-0.276 

-0.385 

-0.018 

.235 

6 

0.447 

2.68 

2.98 

0.447 

0.474 

-0.350 

-0.490 

-0.016 

.335 

7 

0.563 

2.27 

3.76 

0.378 

0.575 

-0.423 

-0.592 

-0.017 

.422 

8 

0.669 

2.00 

4.45 

0.332 

0.648 

-0.479 

-0.671 

-0.023 

.478 

9 

1.000 

2.01 

6.68 

0.333 

0.825 

-0.478 

-0.670 

0.155 

2.260 

10 

1.260 

2.015 

8.40 

0.335 

0.924 

-0.476 

-0.668 

0.256 

2.810 

11 

1.78 

2.025 

11.86 

0.337 

1.074 

-0.472 

-0.660 

0.414 

4.000 

12 

2.51 

2.040 

16.79 

0.339 

1.224 

-0.470 

-0.658 

0.566 

5.680 

13 

2.00 

2.470 

13.35 

0.412 

1.125 

-0.385 

-0.539 

0.586 

5.490 

14 

1.58 

2.940 

10.50 

0.488 

1.022 

-0.312 

-0.437 

0.585 

5.130 

15 

1.24 

3.55 

8.25 

0.590 

0.916 

-0.229 

-0.320 

0.596 

4.860 

16 

1.13 

3.87 

7.52 

0.643 

0.876 

-0.192 

-0.268 

0.608 

4.840 

17 

1.00 

4.23 

6.67 

0.705 

0.824 

-0.152 

-0.213 

0.611 

4.770 

18 

0.89 

4.65 

5.93 

0.775 

0.773 

-0.111 

-0.155 

0.618 

4.590 

19 

0.795 

5.13 

5.30 

0.855 

0.724 

-0.068 

-0.095 

0.629 

4.480 

20 

0.710 

5.64 

4.73 

0.940 

0.674 

-0.027 

-0.038 

0.636 

4.450 

21 

0.60 

5.70 

4.00 

0.950 

0.602 

-0.022 

-0.031 

0.571 

3.800 

22 

0.50 

5.77 

3.33 

0.960 

0.523 

-0.018 

-0.025 

0.498 

3.190 

23 

0.40 

5.83 

2.67 

0.971 

0.426 

-0.012 

-0.017 

0.409 

2.590 

24 

0.30 

5.89 

2.00 

0.980 

0.301 

-0.009 

-0.013 

0.288 

1.960 

Now  if  one  of  the  points,  say  1,  be  taken  as  the  datum  point, 
the  ratios  of  the  temperature  at  various  points  to  the  tem- 
perature of  this  datum  point  and  the  change  of  entropy  from  this 
datum  may  be  figured.  Thus  considering  the  point  2: 


p.Vi 


This  ratio  holds  to  the  point  of  explosion  since  the  unburned 
mixture  after  burning  is  different  in  volume,  due  to  the  change 
on  burning,  and  it  is  necessary  to  change  the  temperature  found 
by  the  above  formula  in  the  following  proportion: 

vol.  of  mixture  • 

vol.  of  burned  gases 


T,  = 


394  HEAT  ENGINEERING 

In  practical  application  the  change  in  volume  due  to  chemical 
action  is  so  slight  that  the  volume  ratio  is  assumed  unity. 

If  the  temperature  of  the  point  1  be  assumed  on  the  T-S 
diagram,  the  height  to  other  points  may  be  found  by  multiply- 

T 

ing  by  the  ratio  7^-     If  the  height  to  1  be  made  1  in.  the  ratio 

1  1 

T 

-^r  will  give  the  heights  to  the  various  points.  Of  course  the 
1  1 

scale  of  temperature  is  not  known  uritil  one  temperature  is 
known.  As  will  be  shown,  much  data  can  be  determined  even 
if  this  scale  is  not  known. 

dv   .        dp 
Now  ds  =  cp  —  -f  cv  — 

S2   —  Si    =    Cp  loge  —  +    Cv   loge  —  (50) 

Vi  PI 

Cp  .  V2     .     ,  PZ  ,_,N 

-?  logic  -  +  logio  jj-  (51) 

Cv  t/i  PI 


S2  —  Si 


AT  Sz  —  Si         S2  —  Si 

Now  -x~5 —  =  - 

2.3  cv        const. 

The  constant  of  this  expression  only  changes  the  scale  of 
entropy  if  the  expression 

is  plotted  as  the  entropy.  Fig.  186  is  obtained  by  this  method 
and  the  area  of  the  figure  which  represents  the  difference  between 
the  heat  added  and  that  taken  away  is  equal  to  the  work  done  or 
the  work  shown  by  the  indicator  card.  If  the  area  scale  of  the 
indicator  card  is  known  in  B.t.u.  per  square  inch  the  area  scale  of 
the  T-S  diagram  will  be  inversely  proportional  to  the  areas  of 
the  two  figures. 

Scale*  =  ^  X  Scalept,  (53) 


Scalep.,,.  = 


cu.  ft.  per  in.  X  Ibs.  per  sq.ft.  per  in. 

778 


Fpv  =  area  of  p-v  diagram 

Fts  =  area  of  T-S  diagram  (54) 

The  scale  fixes  the  heat  and  efficiency  of  the  cycle,  although  the 
component  scales  of  temperature  and  entropy  are  not  known. 
As  soon,  however,  as  one  temperature  is  determined  the  scale 


INTERNAL  COMBUSTION  ENGINES 


395 


of  temperature  will  be  known  and  from  it  and  the  area  scale 
the  entropy  scale  in  B.t.u.  per  degree  can  be  found.  The  area 
abed  is  the  heat  accounted  for  by  the  card.  If  the  curve  be  is 
carried  out  to  e  so  that  abef  is  equal  to  the  heat  per  card  the  area 
dcef  represents  the  reduction  in  the  heat  of  combustion  due  to 


8.00 


7.00 


6.00 


5.00 


4.00 


3.00 


2.00 


1.00 


01 


10; 


9 


22 


d\ 


h 


-0.10     0.0     0.10     0.20    0.30    0.40     0.50    0.60     0.70    0.80    0.90     1.00    1.10    1.20 
Temperature  -Entropy  Diagram  to  Special  Scale 

FIG.  186. — Temperature-entropy  diagram  of  the  gas  engine  card. 

absorption  from  the  cylinder  walls  and  incomplete  combustion. 
The  area  dcgh  represents  the  heat  absorbed  from  the  walls  of 
the  cylinder  during  expansion.  The  area  ibcg  represents  the 
work.  The  theoretical  form  of  the  cycle  should  have  been 
nbem.  abik  is  equal  to  heat  absorbed  by  wall  during  compres- 
sion. 


396 


HEAT  ENGINEERING 


LOGARITHMIC  DIAGRAM 

In  Fig.  187  the  logarithmic  diagram  for  the  indicator  card 
is  drawn  and  from  it  the  values  of  the  exponents  n  are  found. 


S0.50 


0.1 


0.2  0.3  0.4  0.5  0.6 

Logarithm  of  Volume  in  Inches 


0.8 


FIG.  187. — Logarithmic  diagram  of  gas  engine  card. 

TEST  OF  GAS  ENGINE 

As  a  closing  problem  on  gas  engines  the  following  data  is 
taken  from  a  report  on  a  test  of  a  producer  and  two-cylinder  gas 
engine  reported  by  Prof.  H.  W.  Spangler  in  the  Journal  of  the 
Franklin  Institute,  May,  1893.  The  data  is  as  follows: 

2  cylinders  each 14^  in.  X  25  in. 

Coal — by  weight — 

Moisture 4 . 20  per  cent. 

Volatile  matter    (5.80   per  cent.    CH4, 

0.73  per  cent.  H2) 6 .88  per  cent. 

Fixed  carbon 80 . 41  per  cent. 

Ash 8.51  per  cent. 

100.00 

Sulphur 0 . 74  per  cent. 

Time  of  test 525        min. 

Revolutions 84,425  160.76  r.p.m. 

Explosions 81,673  155.44  ex.p.m. 

Mean  gas  pressure lii/fc  in.  =  0.062  Ibs.  per  sq.  in. 

Barometer 14 . 686  Ibs.  per  sq.  in. 

Gas  temperature 75 . 5°  F. 

Room  temperature 81 . 3°  F. 

Jacket  temperature  outlet 99 . 23°  F. 

Jacket  temperature  inlet 62 . 42°  F. 

Exhaust  pyrometer 752 . 6°  F. 

Brake  load 1148.5    Ibs. 

Zero  brake  load 160 . 0 

Brake  arm 3.073  ft. 

B.h.p 92.73 

I.h.p Top  cylinder 64.36 

Bot.  cylinder 66.80     131.16 

Coal..  ..1069.6    Ibs. 


INTERNAL  COMBUSTION  ENGINES 


397 


Fuel  gas  by  volume  CO2. .  . 
02.... 
CO.... 
H2.... 
CH4. . . 
N2.. 


4.02  per  cent. 

0.26 
25.38 

4.51 

1.79 
64.04  100.00 


Exhaust  gas  by  volume 

CO,.. 
CO..' 

lii.-:. 

.   15.60  per  cent. 
.     2.24 
.     0.28 
.   81.93     100.00 

Cooling  water 664  Ib.  in  4  min. 

Gas  for  ignition  tube 840  cu.  ft. 

Relative  humidity  of  air 80  per  cent. 


230  Ibs. 


SOlbs. 


FIG.  188.  —  Average  card  from  test  of  Otto  engine. 


The  following  data  will  be  computed: 

Heat  of  coal       C  =  80.41  X  14,544  = 

H2  =    0.73  X  61,500  = 

CH4  =    5.80  X  24,000  = 


11,690 

449 

1,390 

13,529 


Total  heat  supplied  1069.6  X  13,529  =  14,470,600  B.t.u. 

Carbon  per  pound  of  gas  : 

Relative  carbon  from  CO2  =  4.02  X  12  =  48.24 
Relative  carbon  from  CO  =  25.38  X  12  =  304.56 
Relative  carbon  from  CH4  =  1.79  X  12  =  21.48 


Total  relative  carbon .  .  .   374.3 


Relative  weight  of  gas 
From 


C02  

4.02  X  44  = 

176.5 

O2  

0.26  X  32  = 

8.3 

CO  

25.38  X  28  = 

710.0 

H2 

'  4.51  X    2  = 

9.0 

CH4 

1  79  X  16  = 

28.6 

N2  

64.04  X  28  = 

1792.0 

100.00 


2724.4 


398  HEAT  ENGINEERING 

2724.4 
Hence  molecular  weight  of  mixture  =    —  forT  =  27.24 


Carbon  in  1  Ib.  of  gas  =  -  0.1372 

Carbon  per  pound  coal 

From  C    =  0.8041  X  1  =  0.8041 
From  CH4  =  0.0580  X        - 


Pound  of  gas  per   pound  of  coal  =  Q  ^072  =  6-18  Ibs. 

359 
Volume  of  1  Ib.  of  gas  =  07^4  =  *3.1. 

Volume  of  gas  per  pound  of  coal  =  6.18  X  13.1  =  81.00  cu.  ft. 

Heating  value  of  gas  per  cubic  foot  : 

High  Low 

From    CO  =  0.2538  X    337  =  85.5  85.5 

From     H2  =  0.0451  X    345  =  15.5  13.1 

From  CH4  =  0.0179  X  1071  =     ,  19.2  17.1 


Heat  per  cubic  foot,  B.t.u 120.2  B.t.u.  115.7  B.t.u. 

Heat  in  gas  per  pound  coal  =  81  X  120.2  =9740  B.t.u. 

9740 
'/Efficiency  of  producer  =  1Q_OQ  =  0.72 

(neglecting  temperature) 

=  72%. 
Indicated  work 131.16  X  2546  X  QQ-  =  2,920,000. 

2920000 
Indicated  thermal  efficiency  of  producer  and  engine  =  1 A /t7rtAnn  =  20.2%. 

l'±'±«  UOUU 

20  2 
Indicated  thermal  efficiency  of  engine  =  ^^  =°-281  =  28.1%. 


Delivered  work  =  92.73  X  2546  X  ^jj  =  2,070,000. 

2070000 
Delivered  thermal  efficiency  of  producer  and  engine  =  14470600  =  ^.30  %. 

v/Delivered  thermal  efficiency  of  engine  =      ~  =  0.199  =  19.9%. 


Heat  in  cooling  water  =  664  X  ^  [99.23  -  62.42]  =  3,220,000. 

3220000 
Percentage  of  heat  in  coal  removed  by  jacket  water  =  14470500  = 

v/^ercentage  of  heat  in  gas  removed  by  jacket  water  =  =  30.9  %. 

Air  required  per  cubic  foot  of  gas: 


INTERNAL  COMBUSTION  ENGINES  399 

The  products  of  combustion  for  complete  combustion  and  no  dilution  are : 


Per  cent, 
vol. 

C02 

N2 

H2O 

Air 

OO2  

0.0402 

0.0402 

O2. 

0  0026 

0.0098 

0.0124 

oo  

0.2538 

0.2538 

0.4300 

0.6080 

H2  

0.0451 

0.0451 

0.1070 

CH4  

0  0179 

0.0179 

0.1443 

0  .  0358 

0.1710 

N2 

0  6404 

0  6404 

0.3119 

1.2545 

0.0809 

0.8736 

The  analysis  of  exhaust  gases  shows  15.60  per  cent,  by  volume  of  CO2 
and  0.23  per  cent.  CO.     Now  1  volume  of  CO  would  produce  1  volume  of 

0  23 

CO2.     Hence  if  ^-^  X  0.3119  =  0.0045  cu.  ft.  be  left  as  CO  the  propor- 
tions will  be  as  follows: 

CO2 0.3074 

CO 0.0045 

N2 1.2464 

H2O 0.0809 

Air  Required 0 . 8629 

The  exhaust  gases  contain  2.24  per  cent,  of  O2  and  15.60  per  cent.  CO2 
hence  there  must  be 

2.24 


15.60 

of  O2  per  cubic  foot  of  gas,  or 
0.044 


X  0.3074  =  0.044  cu.  ft. 


0.21 


=  0.21  cu.  ft.  of  air  in  excess. 


This  gives  as  the  total  amount  of  air  per  cubic  foot  0.21  -f  0.8629  = 
1.0739.  The  amount  of  air  theoretically  required  is  0.8629  cu.  ft.  and 
hence  the  per  cent,  excess  is 

24'2  per  cent" 


The  products  of  combustion  without  moisture  will  then  be: 

CO2  ....................  .  ____  .  ........  0.3074  cu.  ft. 

CO  ........  .............  ...  ....  .......  0.0045  cu.  ft. 

N2  .....................  .  ...  ...  ........  1  .2464  cu.  ft. 

H2O  ........................  ,  .......  ..  0.0809  cu.  ft. 

Air  ...................................  Q.2100CU.  ft. 

Total  .....................  .  .........  1.7492cu.  ft. 

These  came  from  1  cu.  ft.  of  gas  and  1.0729  cu.  ft.  of  air. 

The  gas  passes  from  the  wet  scrubber  to  the  engine  and  is  therefore 


400  HEAT  ENGINEERING 

saturated.     Moisture  in  gas  is  sufficient  to  saturate  same  at  75.5°  F.  and  will 
exert  a  pressure  of  0.436  Ib.  and  hence  for  a  barometer  of  14.686  Ibs.  and  a 
pressure  of  0.062  Ibs.  gauge  or  14.748  Ibs.  absolute,  the  moisture  occupying 
1  cu.  ft.  at  0.436  Ib.  would  occupy 
0.436 


14.748 


0.0296  cu.  ft. 


at  atmospheric  pressure  plus  gas  pressure. 

The  air  at  81.3°  F.  is  80  per  cent,  saturated  and  would  exert  a  pressure  of 

0.527  X  0.80  =  0.422  Ibs.  per  sq.  in. 

and  therefore  the  moisture  per  cubic  foot  of  air  under  atmospheric  condi- 
tions would  be 


Now  the  air  and  gas  are  not  under  the  same  conditions  of  pressure  and 
temperature  and  the  1.0739  cu.  ft.  of  air  above  computed  must  be  reduced 
to  the  same  pressure  and  temperature  as  the  atmosphere.  The  pressure  on 
the  gas  is 

14.748  -  0.436  =  14.312  Ibs. 

and  the  air  pressure  is 

14.686  -  0.422  =  14.264  Ibs. 

The  temperatures  are  75.5°  F.  for  the  gas  and  81.3°  F.  for  the  air. 
The  amount  of  air  if  reduced  to  the  conditions  of  atmosphere  will  be  : 


The  mixture  entering  the  engine  with  1  cu.  ft.  of  gas  and  its  molecular 
weight  are  given  below  : 

Gas  ..................................  1.000  cu.  ft. 

Air  .............  ;  .....................  1  .074  cu.  ft. 

H  o  f  gas  ..................  0  .  030  cu.  ft. 

'  \  air  ..................  0.031  cu.  ft. 

Total  ...................  .  ...........  2.135 

The  products  of  combustion  per  cubic  foot  of  gas  and  the  computation 
for  the  mean  molecular  weight  are  given  below: 


CO  

Volume 
0  .  0045  

V  X  mol.  wt. 
...     0.126 

N2  
Air  

H2O. 

1.2464  
0.2100  
f  0.0809) 
\  0  0300  [ 

...   35.000 
...     6.02 

2  67 

CO2  

[  0.0310  J 
0.3074  

.  .  .    13.43 

Total 1.9112 57.246 

Mean  molecular  weight  =  .  QI  10  ==  30.1 


INTERNAL  COMBUSTION  ENGINES  401 

The  values  of  the  specific  heats  for  the  burned  mixture  are  given  below : 

Gas  Value  of  a 

CO 0.0045  X  0.018     =  0.000081 

N2 1.2464  X  0.018     =  0.022500 

Air 0.2100  X  0.018     =  0.003780 

H20 0.1429  X  0.0188  =  0.002690 

CO2 0.3074  X  0.0202  =  0.006280 

0.035331 

Gas  Value  of  b 

CO 0.0045  X  0.185  X  10~5  =  0.0083  X  10~5 

N2 1.2464  X  0.185  X  10~5  =  0.2320  X  10~5 

Air 0.2100  X  0.185  X  10~5  =  0.0388  X  lO"5 

.  H20 0.1429  X  0.668  X  10~5  =  0.1950  X  lO"8 

CO2. .  .  0.3074  X  0.807  X  10~5  =  0.2460  X  10"5 


0.7201  X  10~5 

0.035331 
1.9112 

t  10-5 

-  =  0.3762  X  10~5 


From  this  the  heat  in  the  exhaust  gases  at  752.6°  F.  above  75.5°  F.  is 

fc 


Heat  =  VfCpdt  =  V[a(T2  -  Ti)  +  |(IV  -  TV)]  (55) 


=  1.9112[0.01831(752.6  -  75.5)  +  0-37622X      -(752.6*  -  75.5*)] 

=  1.9112[12.398  -  1.052] 
=  25.70 

Heat  in  exhaust  gases  at  75.5°  F.  due  to  evaporation  of  water: 
Volume  of  H2O  =  0.1429  cu.  ft. 
Weight  of  H,0  -  "-I*  X  14.7  XU4  =  ^^ 

^  X  535.5 
Heat  of  vaporization  =  1049  X  0.0066  =  6.92  B.t.u. 

Total  heat  in  exhaust  gases  per  cu.  ft.  =  32.62  B.t.u. 
Heat  of  combustion  in  carbon  monoxide  in  exhaust  gas 

0.0045  X  337  =  1.518  B.t.u. 

Heat  Balance  per  cu.  ft.  of  gas: 

Heat  supplied 120.20 

Heat  brought  in  above  75.5°  F.  by  air  = 

1.070[81.3  -  75.5]  [0.018] 0.11 

Moisture  in  gas  and  air  =  e^43  x  2.67 1.15 

Total  Heat  Supplied 121.46 

26 


402 


HEAT  ENGINEERING 


Heat  in  indicated  useful  work  (28.1 

per  cent.) 33.8  B.t.u. 

Heat  in  jacket  water  (30.9  per  cent.)  36 . 2 

Heat  in  exhaust  gases { 

\.  D  .  y^ 

Heat  in  CO 1.52 

Difference..                                         .  18.96 


COMBUSTION  OF  FUELS  FOR  BOILERS 


27.8% 

29.8 

26.8 

1.3 

14.3 

100  .~0% 


The  combustion  of  other  fuels  may  be  considered  at  this 
point  with  advantage.  This  combustion  is  practically  the  same 
as  that  of  gases  considered  earlier.  This  union  of  the  elements 
with  oxygen  develops  heat.  The  heat  of  combustion  and  the 
air  required  are  found  in  the  table  on  p.  377  as  was  done  for 
gases.  The  solid  and  liquid  fuels  are  composed  primarily  of 
carbon,  hydrogen,  oxygen  and  sulphur., 

The  solid  fuels  are  coal,  lignite,  peat,  wood  and  certain  waste 
materials  such  as  bagasse.  These  differ  in  their  chemical  com- 
position. This  composition  is  found  by  chemical  analysis. 
When  the  analysis  is  made  in  a  combustion  tube  giving  the 
percentages  of  the  various  constituents,  the  analysis  is  called 
an  ultimate  analysis.  If,  however,  the  analysis  gives  only  the 
moisture,  volatile  matter,  fixed  carbon  and  ash,  it  is  known  as  a 
proximate  analysis.  This  analysis  is  easily  made  and  is  the 
one  made  by  engineers  to  place  a  coal.  The  analysis  is  given 
in  percentages  of  the  various  constituents  as  received  and  on 
a  dry  basis,  eliminating  the  moisture.  This  latter  is  necessary 
to  reduce  the  effect  of  the  variable  moisture  to  zero.  The 
results  of  analyses  are  given  below: 

ANALYSIS  ON  DRY  BASIS 


Ultimate  analysis 

Proximate  analysis 

Heat 
Value 

C 

H 

0 

N 

S 

Ash 

Fixed 
C 

Vol. 
matter 

Ash 

Anthracite  
Semi  Bituminous  
Bituminous  
Lignite 

88.0 
81.0 
75.0 
65.0 
61.0 
51.0 

2.0 
4.8 
5.0 
4.5 
6.0 
7.6 

2.0 
4.0 
9.0 
20.0 
28.10 
4.0 

0.8 
1.5 
1.5 
1.0 
1.9 
37.0 

0.50 
0.75 
1.5 
1.5 

6.7 
7.95 
8.00 
8.00 
3  0 

85 
73 
56 
44 

7.00 
20.00 
36.00 
48.00 

8.00 
7.00 
8.00 
8.00 

13,750 
14,700 
13,800 
10,700 
9,000 
9,000 

Peat  .  . 

Wood  

0  4 

At  times  the  analysis  is  put  on  an  ash-  and  moisture-free 
basis.  This  is  expressed  in  percentages  of  the  constituents 
with  ash  and  moisture  omitted. 


INTERNAL  COMBUSTION  ENGINES  403 

The  constituents  unite  with  the  oxygen  according  to  the 
formulae 

c  +  o2  =  co2. 

H2  +  K02  =  H20. 

s  +  o2  =  so2. 

From  the  atomic  weights  of  the  elements,  the  weights  of  the 
oxygen  required  per  pound  of  carbon,  hydrogen  or  sulphur 
may  be  found,  together  with  the  products  of  combustion.  Thus 
1  Ib.  of  carbon  requires  3%2  IDS-  of  oxygen,  1  Ib.  of  hydrogen  re- 
quires *%  or  8  Ibs.  of  oxygen  and  3%2  Ib.  of  oxygen  are  required 
per  pound  sulphur.  If  these  quantities  are  divided  by  0.232  the 
air  required  per  pound  of  the  various  substances  is  known.  If 
C,  H  and  S  represent  the  parts  of  these  substances  in  1  Ib.  of 
fuel,  the  air  required  is  given  by: 

Lbs.  of  air  per  Ib.  fuel  =  11.5  C  -f  34.5  H  +  4.35  S. 
This  is  given  also  as: 

Lbs.  of  air  =  11.5  C  +  34.5  (H  -  ^)  +  4.35  S. 
or  approximately 

Lbs.  of  air  =  12  C  +  36  #  -- 


The  value  5-  is  substracted  from  H  if  0  is  the  weight  of  the 

o 

oxygen  in  the  fuel,  since  it  is  assumed  that  this  oxygen  is  united 
with  the  proper  amount  of  hydrogen  in  the  form  of  H2O. 
The  heat  of  combustion  has  been  given  by  Dulong. 

Heat  per  pound  =  14,6500+62,100  (H  -  ~^j  (56) 

The  value  given  by  this  formula  is  approximately  that  given  by 
the  bomb  calorimeter.  The  results  of  burning  determined  by  a 
calorimeter  are  given  in  the  table  on  p.  402. 

The  products  of  combustion  of  solid  fuels  are  principally  C02, 
N2  and  excess  air.  If  pure  carbon  is  burned,  the  CO2  formed  is 
just  equal  in  volume  to  the  oxygen  consumed,  so  that  the  maxi- 
mum amount  of  C02  is  21  per  cent,  by  volume.  As  there  is 
some  hydrogen  present  and  as  the  water  occupies  twice  the  vol- 
ume of  oxygen  consumed,  each  per  cent,  of  hydrogen  reduces  the 
percentage  CO2  because,  although  the  water  condenses,  the  nitro- 
gen of  the  air  left  from  its  burning  cuts  down  the  possible  per- 
centage. As  soon  as  oxygen  appears  in  the  burned  gases  it 
indicates  that  excess  air  has  been  supplied.  This  is  advisable  in 


404 


HEAT  ENGINEERING 


many  cases  as  there  is  danger  in  having  some  CO  when  the  air 
is  not  in  excess  and  the  loss  due  to  improper  combustion  is  greater 
than  that  due  to  the  presence  of  excess  air.  This  latter  represents 
a  loss  due  to  a  larger  quantity  of  hot  gas  sent  from  a  boiler  or 
gas  engine.  It  is  shown  that  at  least  a  30  per  cent,  excess  is 
necessary  to  prevent  the  formation  of  CO.  The  curve  of  Fig. 
189  illustrates  the  relation  between  per  cent.  CO2  and  total  air 
supply  as  a  per  cent,  of  the  theoretical  air  supply.  It  is  found 
that  when  the  per  cent.  CO2  is  between  10  and  15  per  cent,  the 
best  results  are  obtained. 

In  analyzing  gases  from  furnaces  a  percentage  of  CO2  of  15 
per  cent,  with  3  or  4  per  cent,  of  O2  represents  good  results. 


35* 
30* 
25* 

|  15* 
10* 


0.8 


0.6o 

•3 


0.4 


50*          0    25*   50*  75*100  125  150  175  200  225  250 
Deficiency  of  Air  Excess  of  Air 

FIG.  189. — Curve  of  relation  between  per  cent.  CO2  and  excess  of  air. 

PROBLEM 

The  analysis  of  the  exhaust  gases  from  a  boiler  gives  the  exact  excess  air. 
Thus  for  a  coal  of  the  following  analysis: 

Per  Cent,  by  Weight 


C... 
H2.. 
02.. 


Ash.. 


and  flue  gases  of  the  following  analysis: 


80.0 
5.0 

10.0 

0.5 

4.5 

100.0 


INTERNAL  COMBUSTION  ENGINES  405 

Per  Cent,  by  Volume 

C02 12.0 

CO 0.5 

O2 5.0 

N2 82.5 

The  amount  of  air  may  be  worked  out.  The  oxygen  and  carbon  in  the  ex- 
haust gas  may  be  found  by  remembering  that  from  Avogadro's  Law,  equal 
volumes  at  the  same  temperature  and  pressure  contain  equal  numbers  of 
molecules.  Hence  each  volume  may  be  considered  to  contain  one  molecule. 
The  weights  of  carbon  and  oxygen  of  the  gas  are  found  as  follows : 

Carbon  from  CO2  =  0.12  X  12  =  1.44 
Carbon  from  CO  =  0.005  X  12  =  0.06 
Total  carbon 1 . 50 

1.44 

Per  cent,  carbon  burned  to  CO2  =  — —  =  96  per  cent. 

1.50 

c\  r\(\ 

Per  cent,  carbon  burned  to  CO  =  — —  =  4  per  cent. 

1.50 

Oxygen  from  CO2  =0.12    X  32  =  3.84 

Oxygen  from  CO    =  0.005  X  16  =  0.08 

Oxygen  from  O2     =0.05     X  32  =  1.60 

Total  oxygen  required 5 . 52 

r    en 

Oxygen  per  Ib.  carbon  =  — — -  =  3.68 

JL  *OU 

1  60 

Pounds  excess  of  oxygen  supplied  per  pound  carbon  =  — —  =  1.067 

1 .50 

Per  cent,  excess  of  oxygen  in  total  oxygen  shown  by  analysis  = 
160  X  100 


552 


29  per  cent. 


The  gas  analysis  does  not  show  the  oxygen  required  for  the  hydrogen  and 
sulphur  and  these  must  be  added  to  that  for  the  carbon  and  from  the  same 
the  oxygen  of  the  coal  is  subtracted.  The  oxygen  required  is  given  by: 

Oxygen  for  carbon  as  burned  and  excess 

oxygen  per  pound  coal' =  0 . 80     X  3 . 68  =  2 . 944 

Oxygen  for  hydrogen =0.05     X8.0  =0. 400 

Oxgen  for  sulphur  to  SO2 =  0 . 005  X  1 . 0  =  0 . 005 

3.349 
Oxygen  in  coal 0 . 100 

•    Net  oxygen  per  pound  coal 3  249 

3.249 

Air  per  pound  coal  =  =  13.95 

0.232 

0  29  X  2  944 
Excess  of  air  in  total  air  supplied  = —  -  =  26.25  per  cent. 


406  HEAT  ENGINEERING 

Products  of  Combustion :        » 

44 

Carbon  dioxide 0.8      X  0.96  X  —  =  2 .810 

\2i 

28 
Carbon  monoxide 0.8     X  0.04  X  —  =  0.075 

Water  vapor 0.05     X  9  =  0.450 

SO2 0.005  X  2  =  0.010 

Excess  air 1.067  X  0.80  X  ~~  =  3.670 

Nitrogen (13.95  -  3.67)  0.768  =  7 . 890 

Ash..,  =0.045 


Total  weight =14.950 

This  checks  the  1  Ib.  of  coal  and  the  13.95  Ibs.  of  air. 

The  moisture  brought  in  by  the  air  may  be  treated  in  the 
same  manner  as  that  used  in  the  problem  of  p.  400.  The 
heat  carried  up  the  stack  may  be  computed  in  a  manner  similar 
to  that  used  in  finding  the  heat  in  the  exhaust  gases  of  a  gas 
engine.  This  quantity  when  divided  by  the  heat  of  combustion 
of  coal  gives  the  percentage  loss  in  the  stack.  This  amounts 
to  about  10  or  15  per  cent,  when  there  is  just  sufficient  air,  while 
with  100  per  cent,  excess  it  amounts  to  20  per  cent.  A  deficiency 
of  air  means  a  loss  due  to  incomplete  combustion.  After  the 
carbon  burns  to  CO2  a  further  passage  through  a  bed  of  hot  coals 
changes  the  C02  to  CO  and  the  water  from  the  burning  of  the 
hydrogen  may  be  dissociated.  On  mixing  the  CO  with  hot  air 
and  having  the  H2  and  O  mix  after  cooling,  in  the  presence  of  a 
flame  the  gases  burn  and  give  out  their  heat  of  combustion. 
With  an  insufficient  or  a  cold  air  supply  above  the  fire  the  CO 
may  not  burn.  If  volatile  hydrocarbons  are  driven  off  in  the 
form  of  smoke  these  must  be  mixed  with  air  in  the  presence  of 
incandescent  fuel  to  prevent  smoke  formation. 

SURFACE  COMBUSTION 

To  cause  gas  to  burn  completely  with  the  theoretical  amount 
of  air  present  or  a  little  above  the  requisite  amount  of  air  the 
method  of  surface  combustion  has  been  suggested.  In  this  a 
mixture  of  gas  and  air  is  delivered  into  a  tube  of  a  burner  and  dis- 
charged into  a  receptacle  filled  with  small  pieces  of  refractory 
material.  If  the  mixture  is  lighted  and  its  velocity  through 
the  tube  is  more  rapid  than  the  speed  of  burning,  the  flame  will 


INTERNAL  COMBUSTION  ENGINES 


407 


not  enter  the  tube  and  cause  an  explosion  but  the  mixture  will 
quietly  burn  and  heat  the  refractory  material  to  brightness  unless 
the  heat  is  removed  by  some  method.  Fig.  190  illustrates  one 
form  of  burner.  In  the  Bone-Schnabel  boiler  this  method  is 
applied,  the  combustion  taking  place  in  the  boiler  flues  which  are 
cither  filled  with  small  pieces  of  a  refractory  substance  or  rods 
of  the  same.  The  heat  of  combustion  then  heats  these  substances 
so  that  the  heat  is  transmitted  to  the  tube  surface  by  radiation 
and  by  conduction  so  readily  that  not  only  is  the  capacity  in- 
creased but  the  efficiency  is  greatly  increased  due  to  the  low 
temperature  of  the  exhaust. 
The  combustion  is  complete. 
In  tests  quoted  in  the  Journal 
of  the  A.S.M.E.  for  Jan., 
1914,  p.  09,  G.  Neuman  re- 
ports a  test  on  one  of  these 
boilers  showing  an  evapora- 
tion of  16.4  to  30.75  Ibs.  per 
square  foot  of  surface  and  an 


FIG. 


190. — Burner  for  surface  com- 
bustion. 


efficiency  of  90  per  cent. 

The  porous  filling  not  only 

causes  the  gas  to  move  at  a  higher  velocity  relative  to  the  tube 
but  these  heated  objects  maintain  combustion  and  thoroughly 
mix  the  gas  and  air  preventing  stratification. 


TOPICS 

Topic  1. — Sketch  the  cycles  of  Beau  de  Rochas,  Atkinson,  Otto,  Lenoir, 
and  Diesel.  Explain  the  action  on  each  line  and  reduce  expressions  for  the 
thermal  efficiency  of  each  cycle. 

Topic  2. — From  the  efficiency  of  the  Otto  cycle  on  the  air  standard 


773  =  1  - 


Tz-Tb 
Td-  To 


Reduce  the  expressions 


Th 
_ 


FA  *-i 


1  + 


Give  reasons  for  the  impossibility  of  reducing  the  expressions  for  efficiency 
of  the  other  cycles  by  this  method.  What  is  meant  by  air  standard? 

Topic  3. — Explain  the  action  of  a  four-cycle  gas  engine  and  a  two-cycle  oil 
engine.  Draw  cards  from  each.  Explain  how  these  engines  are  governed. 
Explain  the  methods  of  ignition. 


408  HEAT  ENGINEERING 

Topic  4. — What  affects  the  rapidity  of  combustion?  Draw  an  indicator 
card  showing  slow  burning.  Sketch  a  curve  showing  the  variation  of  gas 
temperature  in  a  four-cycle  gas  engine.  Is  there  much  cyclic  change  in  the 
metal  of  the  cylinder  walls? 

Topic  5. — Mention  the  various  fuels  used  in  internal  combustion  engines. 
Give  some  of  the  peculiarities  of  each.  What  are  gas  producers?  What  are 
the  two  general  types?  Sketch  one  of  them. 

Topic  6. — Derive  the  expressions  for  finding  the  amount  of  oxygen  and  air 
to  burn  a  cubic  foot  and  a  pound  of  C2H4  together  with  the  expressions  for 
the  products  of  combustion.  Show  how  to  find  the  heat  per  cubic  foot  if 
the  heat  per  pound  is  21,400  B.t.u.  Find  the  specific  heats  per  cubic  foot  if 
cp  =  0.404  andc,,  =  0.340.  On  what  laws  are  these  based? 

Topic  7. — Derive  the  formulae 

B.    1544 


Mol.  wt., 


mol.  wt. 

2  mol.  wt.  X  vol. 
2  vol.  ~ 


Topic  8. — Derive  the  formulae  for  finding  the  temperatures  at  the  corner 
of  the  Otto  cycle  assuming  the  value  of  cv  to  be  a  constant. 

Topic  9. — Derive  the  equation  for  the  adiabatic  of  a  gas  when  cv  —  a'  + 
bT.  Derive  the  equation  for  n  on  this  line  in  terms  of  the  temperatures. 

Topic  10. — Explain  the  method  of  finding  temperature  at  the  ends  of 
compression  and  expansion. 

Topic  11. — Explain  by  formulae  the  method  of  finding  the  temperature 
after  explosion. 

Topic  12. — Explain  the  construction  of  the  T-s  diagram  for  the  gas  engine 
deriving  the  expressions  for  the  construction  of  the  figure. 

Topic  13. — Explain  the  construction  of  the  logarithmic  diagram  of  the 
gas-engine  card.  What  data  can  be  found  from  this  card ?  What  data  must 
be  known  to  construct  the  card? 

Topic  14. — Explain  the  method  of  finding  the  amount  of  dilution  from  the 
analyses  of  the  fuel  gas  and  exhaust  gas. 

Topic  15. — Explain  the  method  of  finding  the  efficiency  of  a  producer. 

Topic  16. — Explain  the  method  of  finding  the  heat  loss  in  the  exhaust  gases 
and  in  the  jacket  water. 

Topic  17. — Explain  the  method  of  finding  the  heating  value  of  a  gas  from 
it  chemical  composition.  Explain  how  to  find -the  heat  equivalent  of  the 
work. 

Topic  18. — Explain  how  to  find  the  amount  of  air  per  pound  of  fuel  in  a 
boiler  test  from  the  coal  analysis  and  gas  analysis.  What  is  Dulong's 
formula?  What  is  surface  combustion?  Explain  the  construction  of  the 
Bone-Schnabel  boiler. 

PROBLEMS 

Problem  1. — The  clearance  of  an  engine  is  30  per  cent,  of  its  longest  stroke. 
The  initial  temperature  is  150°  F.  The  heat  per  cubic  foot  of  mixture 


INTERNAL  COMBUSTION  ENGINES 


409 


is  70  B.t.u.  The  radiation  loss  during  explosion  is  35  per  cent.  Find  the 
temperatures  at  the  corners  of  Otto  cycle  and  Atkinson  cycle  assuming  the  air 
standard.  Find  the  Carnot  efficiency  of  each,  the  theoretical  efficiency  and 
the  type  efficiency. 

Problem  2. — A  Diesel  engine  has  a  final  pressure  of  600  Ibs.  per  square 
inch.  What  is  the  clearance  to  give  this  from  atmospheric  pressure  at  the 
beginning  of  the  stroke  (n  =  1.38).  If  the  original  volume  including 
clearance  is  1  cu.  ft.  of  air  at  160°  F.,  find  the  weight  of  crude  oil  given  in 
this  chapter  which  could  be  burned  by  this  air  with  25  per  cent,  dilution. 
If  75  per  cent,  of  the  heat  is  available  find  the  temperature  and  volume 
after  burning,  using  the  air  standard. 

Problem  3. — Find  the  products  of  combustion  of  1  Ib.  of  crude  oil  burned 
in  a  Diesel  engine  with  15  per  cent,  dilution.  Find  the  temperature  resulting 
therefrom  with  a  25  per  cent,  loss  to  the  jacket  considering  the  actual  com- 
position of  the  gas  and  .assuring  the  pressure  constant. 

Problem  4. — Find  the  temperature  TI  if  the  clearance  is  30  per  cent,  and 
the  gas  and  air  outside  has  a  temperature  of  60°  F.  with  cv  per  cubic  foot 
0  014  and  the  heat  per  cubic  foot  78  B.t.u.  Find  the  other  temperatures 
assuming  20  per  cent,  loss  to  jacket. 

Problem  5. — Assume  a  blast-furnace  gas  of  form  given  in  this  chapter 
mixed  with  10  per  cent,  excess  air.  Find  the  cubic  feet  of  air  per  cubic 
foot  of  gas.  Find  the  value  of  T2  considering  the  actual  composition  of 
the  mixture  if  T\  =  600°  abs.  and  the  clearance  is  25  per  cent.  Find  T3 
with  10  per  cent.  loss.  Find  T^. 

Problem  6. — Sketch  an  indicator  card  from  an  engine  with  25  per  cent, 
clearance  and  draw  the  T-s  diagram  and  logarithmic  diagram.  Find  the 
heat  added  if  the  actual  efficiency  is  28  per  cent.  Find  the  scales  of  the 
figure  if  TI  is  found  to  be  125°  F.  What  is  the  maximum  value  of  jPi. 

Problem  7. — The  following  results  are  obtained  from  a  test: 


Coal 

C 80.0  per  cent. 

CH2 6.0  per  cent. 

H2 0.5  per  cent. 

Ash 10 . 0  per  cent. 

Moisture    3 . 5  per  cent. 


Producer  Gas 


Exhaust  Gas 


C02.. 

...  4.0  per  cent. 

Co2... 

...15.8  per  cent. 

02... 

.  .  .  0.5  per  cent. 

02.... 

.  .  .  2.5  per  cent- 

CO.. 

.  .  .25.5  per  cent. 

CO... 

...  0.2  per  cent- 

H2... 

...  5.0  per  cent. 

N,.... 

...81.5  per  cent- 

CH4. 

.  .  1.8  per  cent. 

N2 63.2  per  cent. 


100.0 


100.0 


100.0 


Time  of  test 1,200  min. 

Revolutions 240,000 

Explosions 110,000 


F. 


Gas  pressure 

Barometer 

Gas  temperature. . . 
Air  temperature .  .  . 
Relative  humidity  . 


Exhaust  temperature.        800C 
Jacket  temperature  70° 

Jacket  temperature. .  .         120° 

2  in.  water     Coal 4,000  Ibs. 

29.9  in.  Gas 320,000  cu.  ft. 

80°  F.  I.h.p 200 

70°  F.  B.h.p v. ...         160 

60  per  cent.     Cooling  water. ..!....  178,000  Ibs. 


Compute  the  various  losses  and  efficiencies  and  make  a  heat  balance. 
Problem  8. — The  heating  value  of  1  Ib.  of  coal  as  fired  is  14,680  B.t.u. 


410  HEAT  ENGINEERING 

The  equivalent  evaporation  from  and  at  212°  F.  per  pound  of  coal  is  11.10 
Ibs.     Find  the  efficiency  of  the  boiler. 

Problem  9.  —  In  Problem  8  the  following  was  found  :  The  flue  gas  analysis 
by  volume  gave  0.18  per  cent.  CO,  6.49  per  cent.  CO2,  13.09  per  cent.  O2, 
and  the  coal  analysis  by  weight  was  as  follows  : 

Moisture  ......................    1.0  per  cent. 

Ash  ...........................   6.9 

Hydrogen  .....................   2.8 

Oxygen  ........................   4.2 

Carbon  ........................  83.0 

Sulphur  ........................  0.8 

Nitrogen   ......................    1.3 

100.0 

Find  the  amount  of  air  per  pound  of  coal.  Find  the  excess  air.  Find  the 
heat  value  by  Dulong's  formula  and  check  with  value  by  calorimeter  given 
in  Problem  8. 

Problem  10.  —  In  Problem  9  find  the  composition  of  the  exhaust  gases  per 
pound  of  coal  assuming  that  the  air  at  a  temperature  of  75°  F.  has  a  wet 
bulb  temperature  of  67°  F. 

Problem  11.  —  With  the  results  of  Problem  10  find  the  heat  carried  away  in 
the  flue  gases  due  to  a  temperature  of  427°  F.  and  the  heat  available  from  the 
CO  present.  Express  these  as  percentages  of  14,680  B.t.u. 

Problem  12.  —  With  coal  at  $4.40  per  ton  what  is  the  cost  of  producing 
1000  Ibs.  of  equivalent  evaporation,  using  data  of  Problem  8.  Equivalent 
evaporation  from  and  at  212°  F.  is  equal  to  the  actual  evaporation  multiplied 
by  the  factor  of  evaporation.  Factor  of  evaporation  is  the  amount  of  steam 
evaporated  from  and  at  212°  F.  by  the  same  amount  of  heat  as  will  evaporate 
1  Ib.  of  water  at  the  given  feed  temperature  into  steam  at  the  given 
condition  for  the  boiler.  This  is  given 


f  = 


r212 


In  this  problem  find  the  factor  of  evaporation  for  a  gauge  steam  pressure 
of  121.3  with  the  barometer  of  29  in.  if  the  feed  temperature  is  105°  F.  and  the 
quality  of  the  steam  is  65°  F.  superheat.  Find  the  cost  of  producing  1000  Ibs. 
of  actual  steam. 


CHAPTER  X 
REFRIGERATION 

AIR  MACHINES 

The  refrigerating  machine  operates  by  placing  a  substance 
in  such  a  condition  that  its  temperature  is  above  the  tempera- 
ture of  a  water  supply,  and  after  the  removal  of  heat  from  the 
substance  by  this  supply  the  condition  of  the  substance 
is  so  changed  that  it  will  abstract  heat  frpm  a  body  of 


FIG.   191. — Air  refrigerating  machine. 

low  temperature.  The  temperature  of  the  substance  is  changed 
by  changing  the  pressure  on  the  substance.  This  is  accomplished 
by  machines  in  which  a  gas  such  as  air  or  a  vapor  such  as  ammonia 
or  carbon  dioxide  is  used.  The  air  machine  is  shown  in  Fig. 
191.  Air  is  compressed  in  the  cylinder  A  from  a  pressure  pi 
to  a  pressure  p^.  At  the  pressure  p2  it  is  discharged  through  a 

411 


412 


HEAT  ENGINEERING 


pipe  system  B  which  is  cooled  by  water  from  the  supply  C.  In 
this  case  the  cooling  of  the  air  reduces  its  volume  so  that  when  this 
air  is  taken  to  a  second  cylinder  D  it  occupies  less  space.  The  air 
is  then  allowed  to  expand  in  the  cylinder  D  to  the  original  pres- 
sure, doing  work  at  expense  of  its  intrinsic  energy,  and  hence  its 
temperature  falls  so  much  that  on  entering  the  coil  E  in  the  tank 
or  room  F,  this  air  will  remove  heat  from  the  brine  in  the  tank  or 
from  the  air,  if  placed  in  a  room.  This  air  is  taken  back  to  the 
compressor  after  removing  heat  from  the  room  F.  The  cycle  is 
shown  in  Fig.  192.  At  times  the  air  is  discharged  into  the  room 
F  instead  of  passing  through  the  coil  and  air  is  sucked  from  the 

room.  The  system  is  then  called  an 
open  system  to  distinguish  it  from 
that  of  Fig.  191  which  is  called  a 
closed  system. 

If  the  temperature  of  the  air 
sucked  from  the  cool  room  F  is  TI 
and  the  pressure  is  pi,  the  condition 
of  1  Ib.  of  air  is  shown  by  the  point  1 ; 
the  volume  4-1  is  sucked  in  on  the 
line  4-1  if  clearance  is  neglected. 
This  air  is  compressed  along  the  line 
1-2  and  the  line  is  practically  an 
adiabatic.  The  temperature  T2 
given  by 


Compressor 


IS 


m  m    /F2 

1  2    =    ill 


k  -  1 
k 


(1) 


Combined  Cord 


FIG.  192.— Cycle  of  the 
machine. 


air 


This  air  occupies  the  volume  3-2 
and  the  card  4123  represents  the 
work  done  on  the  compressor  A. 
The  air  is  forced  out  and  is  cooled 
off  in  the  cooler  B  to  a  temperature 
T$  fixed  by  the  temperature  of  the 

water  at  the  point  of  discharge  of  the  air.  In  a  countercurrent 
cooler  this  may  be  less  than  the  temperature  of  the  outlet 
water.  It  is  about  10°  F.  above  the  temperature  of  the  water 
at  the  point.  When  this  air  passes  into  the  expansion  cylinder 
its  volume  will  be  shown  by  the  point  5.  The  admission  line 
is  3-5.  The  air  expands  to  the  lower  pressure  and  its  condi- 
tion is  shown  by  point  6.  This  action  is  also  adiabatic,  so  that 


REFRIGERATION  413 


n  ==  Tl 

Ts  =  Tb~  (4) 

It  is  thus  seen  that  TI  is  fixed  by  the  temperature  of  the  refrig- 
erator, being  the  room  temperature  in  the  case  of  an  open  system 
or  10°  to  15°  F.  less  for  a  closed  system;  while  T5  is  fixed  by  the 
temperature  of  the  cooling  water,  T2  and  T6  are  fixed  by  the  pres- 
sure ratios  after  TI  and  T5  are  found. 

The  work  done  on  the  compressor  is  4123  while  that  done  by 
the  air  in  the  motor  is  3564.  The  difference  or  1256  shown  by 
combining  the  cards  is  the  net  work  required.  This  work  is 
also  the  difference  between  the  heat  on  the  top  and  bottom  lines 
since  there  is  no  heat  on  the  curved  lines. 

AW  =  cp  {[T2  -  T6]  -  [Ti  -  T6]}  (5) 

It  is  seen  that  heat  removed  by  the  cooler  is  equal  to  the  heat 
removed  from  the  refrigerator  plus  the  work  done  on  the 
substance. 

But  the  heat  to  change  the  volume  from  2  to  5  is  the  heat  re- 
moved by  the  cooling  water  while  that  added  from  6  to  1  is 
that  taken  from  the  refrigerator.  Hence 

Qref.    =    Cp[T,   -    T,]  (6) 

and  Qcooler  =  cp[T2  -  T6]  (7) 

The  efficiency  or  better  the  refrigerative  performance  is  the 
amount  of  refrigeration  per  unit  of  work.  This  is  given  by 

Vr   =  f  rrri      _   m  i          TrF  T~\\  (8) 

C»U.li—  i  sj  —  [1 1  —  1  ejj 
_  1 

tyr    —    rp     _   rp 


i  rfi  rti  rji 


fji       /TT      '       nn     '       rji 

rji  fji 

since  =r  =  TTT 

1  6  1  I 

,  L  5  I  Q 

K,"  f! 


414  HEAT  ENGINEERING 

Hence 


(9) 


T 


These  expressions  are  equal  to  the  lower  temperature  on  one 
of  the  adiabatics  divided  by  the  difference  in  temperature  on 
that  adiabatic.  The  expression  is  greater  than  unity. 

The  refrigeration  produced  by  any  machine  is  usually  measured 
in  tons  of  refrigeration  per  24  hrs.  The  amount  of  heat  liberated 
when  1  Ib.  of  ice  melts  is  144  B.t.u.  (best  value  143.5  B.t.u.), 
and  hence  the  heat  equivalent  to  1  ton  of  refrigeration  is  288,000 
B.t.u.  in  24  hrs.  which  is  200  B.t.u.  per  minute.  The  number  of 
pounds  of  air  required  for  this  refrigeration  is 

200 
Lbs.  of  air  per  min.  per  ton  in  24  hrs.  =  —  ^  ---  ^T  =  Ma  (11) 

Cp[l  i  —   1  6J 

The  amount  of  cooling  by  the  air  cooler  is 

Qc  =  Macp[T2  -  T6]  (12) 

The  water  required  for  this  is  given  by 

Macp[Ts  -  T,] 
3'o  -fl'« 

Mw  =  weight  of  cooling  water  per  min.  per  ton  in  24  hrs. 
q'0  =  heat  of  liquid  of  cooling  water  at  outlet  temperature  t0 
q'i  =  heat  of  liquid  of  cooling  water  at  inlet  temperature  t{. 

The  horse-power  required  for  the  production  of  1  ton  per  24 
hrs.  is 


h.p.  =  Macp(T,  -  Tb-  (T,  -  T6)]  (14) 

Qr         778 


rjr        33000 

0 

(16) 


The  volume  of  the  compressor  is  Vi  while  that  for  the  ex- 
pansion cylinder  is  F6.     These  are  each  given  by 

MgBT,  MaBTl 


~^PlXcl.  factor 
MaBT6 


. 


_ 
?>e  X  cl.  factor 


REFRIGERATION  415 

The  above  equations  are  computed  for  cylinders  without 
clearance  and  friction,  for  non-conducting  cylinder  walls  and 
for  dry  air.  These  three  things  affect  the  results  of  computa- 
tions, but  because  their  effects  are  seen  by  comparison  with 
perfect  conditions  the  equations  for  perfect  conditions  are  given 
although  not  used  in  practice. 

PROBLEM 

Assume  that  a  3-ton  machine  was  desired  to  operate  between  14.7  Ibs. 
abs.  and  58.8  Ibs.  abs.  and  that  the  temperatures  of  the  cooling  water  were 
50°  F.  to  65°  F.  and  that  for  the  cool  room  was  25°  F.  It  will  be  assumed 
further  that  the  air  from  the  expansion  cylinder  is  discharged  into  the  room 
to  be  -refrigerated.  From  the  above  the  following  is  true: 

Tl  =  25°  F.  =  485°  abs. 

T5  =  50°  +  15°  =  65°  F.  =  525°  abs. 

(assuming  a  counter-current  flow) 

/KgQ\  <^1 

Tz  =  485  T   1A  =  485  X  1.486  =  720°  abs.  =  260°  F. 


T6  =  525  X          =  354°  abs.  =    -  106°  F. 

485 

*p  =  720  -  485  =  2'06 
Qr  =  200  X  3  =  600  B.t.u.  per  min. 

600 
Ma.  =  Q.24[485  -  354]  =  19'08  lbs<  of  air  per  min> 

=  19.08  X  0.24[720  -  525]  =  892  B.t.u.  per  min. 


778 
H.p.  =  (892  -  600)  —       =  6.86 


600  778 

also  H.p.  =  2^3   X  33^  =  6.85 

The  water  required  is 

892 
Mv,  =  33  i  _  i  g  i   =  59-4  Ibs.  per  min. 

Displacements  per  minute  are  : 

T.  19.08  X  53.35  X  485       0_r 

Vcomp.  =  14  7  x  144  --  =          C  per  mm' 

v  19.08  X  53.35  X  354 

Vexp.  =  14.7  x  144  =          CU'        per  mm' 

The  machine  considered  in  the  problem  above  is  one  in  which 
the  discharge  has  been  into  an  open  space.  Such  a  system  may 
be  called  an  open  system.  If  now  the  discharge  takes  place 
in  a  closed  pipe  line  (the  closed  system)  it  would  be  possible 
to  have  the  pressure  pi  higher  than  atmospheric  pressure.  Sup- 
pose that  pi  is  made  58.8  Ibs.  abs.  and  p2,  235.2  Ibs.  abs.  so 


416  HEAT  ENGINEERING 

that  the  ratio  —  is  the  same  as  before.     If  under  these  condi- 

Pi 

tions  the  above  problem  is  computed  the  results  will  be  the  same 
except  the  displacements  which  will  be  one-fourth  of  their  former 
values.  This  is  due  to  the  higher  pressures.  This  "dense  air 
machine,"  as  it  is  called,  is  used  because  it  decreases  the  dis- 
placement of  the  cylinders  and  hence  the  size  or  speed  of  the 
machine.  This  is  the  only  reason  for  its  use. 

The  problems  must  now  be  investigated  for  clearance  and 
friction.  Since  friction  increases  the  work  of  the  compressor 
and  decreases  the  work  obtained  from  the  expansion  cylinder, 
the  work  done  by  each  of  these  machines  must  be  considered 
separately,  instead  of  obtaining  the  net  work  1256  as  before. 

The  work  of  the  compressor  1234  is  the  same  with  or  without 
clearance.  It  is  equal  to 

If      (19) 

MB[Tl  -  Til 


If  n  =  k  this  becomes 

Wcomp  =  MJc^T,  -  T2]  (20) 

To  make  this  work  as  small  as  possible,  jackets  may  be  used 
on  the  cylinder  changing  k  to  n  and  making  n  as  low  as  1.38. 


FIG.  193. — Effect  of  friction  in  air  machines. 

If  the  friction  fraction  is  /,  the  work  must  be  multiplied  by 
(1  +  7)  or 

1'2'34  =  work  of  compression  =  (!  +  /)      n    MB[Tl  -  T2]  (21) 


For  the  expansion  cylinder  the  work  obtained  is  (1  —  /')  times 
the  work  without  friction.     This  work  is  independent  of  the 


REFRIGERATION  417 

clearance  if  the  expansion  and  compression  are  each  complete. 
The  work  then  becomes : 

35'6'4  =  work  of  expansion  =  (1  -  f')pQV6  g-z~i[1~  (~)~*~ 
or  work  =  (1  -  f)MJcp[T5  -  T6]  (22) 

The  value  of  n  in  the  expansion  cylinder  is  k  as  it  is  desired 
to  have  as  low  a  value  of  TQ  as  possible  and  this  is  accomplished 
by  having  as  large  a  value  of  n  as  possible.  By  lagging  the 
expansion  cylinder  and  omitting  any  jacket  this  n  is  made  k 
as  no  heat  is  taken  away. 

The  difference  of  the  two  expressions  for  work  will  give  the 
work  required  from  the  outside  to  drive  the  machine. 

Of  course  with  n  on  compression  (1.38)  different  from  that  on 
expansion  (1.4)  the  equality  of  cross  products 


is  not  true.     772  and  T6  must  each  be  found. 

/r>  \  n~  * 

r.  -  Ti(?)T  (23) 


(24) 

The  effect  of  clearance  air  on  the  temperatures  is  important 
to  consider.  The  air  left  in  the  cylinder  at  the  end  of  compres- 
sion is  at  a  temperature  T2.  When  this  expands  down  to  atmos- 
pheric pressure  the  temperature  is  TI  so  that  the  fresh  air  enter- 
ing at  this  temperature  mixes  with  it.  In  the  expander  the 
air  in  the  clearance  space  at  temperature  T§  is  compressed  to 
the  temperature  T5  by  complete  compression  and  so  mixes  with 
air  from  the  cooler  at  that  temperature.  If  this  compression 
were  not  complete  this  temperature  T$  would  not  be  reached 
and  the  effect  of  this  would  be  to  give  a  different  temperature 
of  the  mixture  at  the  beginning  of  expansion.  This  incomplete 
compression  would  require  an  excess  amount  of  air  from  the 
supply  to  fill  the  clearance  space.  This  free  expansion  into  the 
clearance  space  would  warm  the  air  in  the  cylinder  and  produce 
a  higher  temperature  at  the  beginning  and  at  the  end  of  expan- 
sion. It  would  mean  a  decrease  in  refrigerating  effect  and  would 
also  decrease  the  work  obtained  from  the  expander  per  pound  of 
air  used. 

27 


418  HEAT  ENGINEERING 

If  the  expansion  were  incomplete  the  temperature  at  the  end 
of  expansion  would  not  be  as  low  as  T&  due  to  the  pressure  range 

—  being  less  than  —  or  —  and  as  a  result  the  temperature  at 
Pe'  Pe  Pi 

the  end  of  free  expansion  or  the  discharge  temperature  would  be 
higher  than  T&.  This  reduces  the  refrigerating  effect.  The 
work  obtained  from  a  given  amount  of  air  has  been  decreased 
by  the  incomplete  expansion  and  compression.  For  these  two 
reasons  the  performance  is  decreased  by  a  very  material  amount 
and  the  endeavor  is  made  to  have  complete  expansion  and 
compression. 

To  eliminate  the  effect  of  clearance  the  compression  in  the 
expansion  cylinder  is  such  that  the  pressure  at  the  end  of  com- 
pression is  equal  to  pressure  of  the  compressor  and  the  expansion 
is  carried  down  to  the  back  pressure.  The  action  is  then  equiva- 
lent to  a  cylinder  without  clearance. 

The  moisture  in  the  air  has  an  effect  on  the  efficiency  of  air 
refrigerating  machines.  The  compression  of  moist  air  may  cause 
some  of  the  moisture  to  condense  giving  out  heat  and  warming 
the  air  while  a  further  condensation  and  even  freezing  in  the  ex- 
pansion cylinder  causes  the  liberation  of  additional  heat  reducing 
the  refrigerating  effect.  Air  will  be  removed  in  short  time. 

Suppose  that  air  of  relative  humidity  p,  temperature  T\  and 
pressure  pi  occupies  the  volume  V\.  The  weight  of  the  air  is 


pti  =  sat.  steam  pressure  corresponding  to 
The  weight  of  the  moisture  Mm  is 


m\  =  weight  of  1  cu.  ft.  of  steam  at  temperature 
This  may  also  be  approximated  by  the  formula 


Mm  =  fmlVl  (26) 


(27) 


t.  steam  ~     18 
For  the  complete  mixture  of  air  and  moisture 

(Ma  +  Mm)BmxT=pV  (28) 


1544  1544 


REFRIGERATION  419 

After  compression  to  the  pressure  pz  the  volume  becomes 

.  (29) 


and  7T2=771  (30) 

Now  -y^  =  weight  of  moisture  in  1  cu.  ft.  (31) 

hence,  if 

ra2  =  weight  of  1  cu.  ft.  of  saturated  steam 

Mm 

Y 

-  —  Pz  =  relative  humidity  after  compression          (32) 

1Yl<i 

p2  is  usually  less  than  unity  because  of  the  high  temperature 
at  the  end  of  adiabatic  compression. 

The  work  of  compression  of  the  mixture  of  air  and  vapor  is 
considered  to  be  that  of  a  perfect  gas  as  the  steam  in  the  air  is 
superheated.  If  adiabatic  this  becomes  : 


=  j  [Maca  +  MmCm](T2  -  TJ  (33) 

ca  =  specific  heat  at  constant  pressure  for  air. 

cm  =  specific  heat  at  constant  pressure  for  moisture. 

1         1544      1.3  _ 
cm  -          X  X         -   0.48. 


If  this  air  is  cooled  in  the  cooling  coil  to  a  temperature  T$, 
the  moisture  may  or  may  not  be  sufficient  to  saturate  the  air. 
To  find  out  the  amount,  it  is  known  that 

Vs  =  V2  ^  (34) 

•L  2 

Mrn 

and  ^3  /Q  .. 

—  -  Pa  (35) 

M 

If  m3  <  -^  (36) 

Mm  —  mzV'z  =  amount  condensed. 


If  the  moisture  condensed  is  removed  by  a  separator,  the 
weight  of  the  material  left  is 


420  HEAT  ENGINEERING 

Ma  +  m3F'3  (37) 

in  place  of  Ma  +  Mm 

The  volume  of  this  is     7'3  =    MaBaT*  (38) 


The  value  of  Qc  is 

Qc  =  Macp[T*--T<i]-{-Mmcpm[T2  -  T3]  +  [Mm  -  m>V'9]r9      (39) 

The  moisture  which  enters  the  expansion  cylinder  is  practi- 
cally all  condensed  and  frozen  by  the  cool  walls  of  the  cylinder 
liberating  heat  represented  by  C. 

C  =  m3Vfs[cp(T2  -  r32)  +   r32  +  144]  (40) 

This  is  gradually  restored  during  expansion. 

The  volume  of  the  air  remaining  is 

P* 

The  air  and  ice  in  the  cylinder  now  expand  with  the  gradual 
return  of  the  heat  C  from  that  produced  by  the  ice  formation 
giving: 

Macvadt  +  (m3V's)cidt  +  Apdv  =  +  dC  (42) 

Macva  +  m3F/3c,  dt  dV  _dC_ 

B  T^       a      V    ~  BT 

Q 

assume  dC  =  —  ^  -  ^-  dt 

1  3  —  1  4 

'  Ts  Vs  C  T3 

T    +  MaA   log,      r  =     - 


Z?       (P* 
V,  =: 
C1  T 

[MaCva  +  WaV'sCt  +   rr  rn     +   MaAB]  loge   ^ 

i  3  —    1  4  J  4 


(43) 


NOW  Cva  +  -4.^    =    C 

Hence 


T*  =  MaAB  loge  S     (44) 

This  is  solved  for  7%  by  trial,  knowing  all  the  other  terms.     If 
z  —  Tt  is  assumed  from  an  approximate  value  of  T4  from 


(45) 

the  equation  leads  directly  to  T± 


REFRIGERA  TION  421 

The  value  of  T±  then  gives  the  heat  taken  from  the  refrigerator 

G    —   M  r  (T A  •-  7M  M-fi"! 

r    —    •'•'-'-  a^p  \  •*•  4  •*•  I/  *%•*'•'/' 

The  moisture  is  not  considered  at  this  point  as  it  has  been 
condensed  and  frozen  in  the  expansion  cylinder  and  does  not 
enter  the  refrigerator. 

The  work  done  in  the  expansion  cylinder  is  best  found  by  the 
following : 


rm 

pdv  — 

tJ   V3 

C        (*T* 

— L* 


We  =  p3V3  +|    pdv  -  piV*  (47) 

From  (42): 


'  C 

M*"+'m'V*«  +  jr-=T<\    dt  (48) 


Macva  +  mzV'zd  +  y  _  TJ  [T3  -  T,} 

Since  the  moisture   has   been    eliminated   on  the   expansion 
curve  p3V3  -  p^V4  =  MaB  [T3  -  ITJ  (49) 


and  j5+=  (50) 


Macp  +  m3y3ci  +  Ts  _  y    [Ts  -  T,]        (51) 

The  work  required  from  the  outside  is 

(1  +  f)W.  -  (1  -  f)W.  (52) 

and  the  refrigerating  effect  is 

Qr  =  Macp  (Tl  -  T,)  (53) 

To  apply  the  above  formulae  assume  that  the  air  enters  the  compression 
cylinder  at  25°  F.,  with  a  relative  humidity  of  80  per  cent,  and  is  compressed 
to  4  atmospheres.     The  water  in  the  cooler  is  sufficient  to  cool  this  to  65° 
F.     Assume  that  Vi  =  1  cu.  ft. 
p25°  =  0.065 
Pp  =  0.065  X  0.8  =  0.052 

[14.7  -  0.052]  X  144  X  1 
M°  =  -          53.34X485"      ~  =  0.082  Ibs. 
Mw  =  0.8  X  0.00023  =  0.00018;  call  this  0.0002 


This  is  a  check. 
4=  0.372 

0.4 

=  485  X  (4)1-4  =  720°  abs.  =  260°  F. 

0.0002 

0.372 
=  0086" 


422  HEAT  ENGINEERING 

or  there  is  not  enough  moisture  to  saturate  the  air.  The  air  is  now  cooled 
to  65°  F.  as  in  the  problem  on  page  415. 

525 
73  =  0.372  X  720  =  0.272 

0.0002 

0.272      _  0.000753 
p3  ~  0.000979  ~  0.000979  ~ 

Hence  none  of  the  moisture  is  condensed  in  the  cooling  coil.  If  p3  were 
greater  than  unity  there  would  be  some  condensation.  After  the  determina- 
tion of  the  amount  of  condensation  a  recalculation  of  Vs  would  be  necessary 
as  some  of  the  water  has  been  removed  from  the  mixture  leaving  the  air 
saturated. 

The  moisture  having  a  specific  weight  of  0.00073  would  be  saturated  at 
56°  F.  so  that  this  moisture  is  superheated  9°  F. 

Approximate  value  of  TV 

k-l  QA   ' 

Tt  =  T3  (-        k   =  525  X  ()L4=  354°  abs.  =    -  106°  F. 


C  =  0.0002[(65  -  32)  ^  +  1071.7  +  144]  =  0.2462 


. 

0.082  X  0/24  +  0.0002  X  0.5  +  ^r^J  ^  log*       =  -^  log,  4 

0.02119  ,        Ts       0.082  . 
~53^T  loge  Y*  =  ~778  loge  4 

log        =  °-161  =  loS  L45 


T*=  j        =  361°  abs.  =    -  99°  F. 
We  =  778[0.02119][65  -  (  -  99)]  =  2695  ft.-lbs. 

0.4 
Wc  =  ^1  X  14'7  X  144  X  1  [l  -  (4)  1<AJ  =    -  3600  ft.-lbs. 

Net  work  without  friction  =  3600  -  2695  =  905  ft.-lbs. 

Net  work  with  friction  =  3600  X  1.10  -  2695  X  0.9  =  1534  ft.-lbs. 

Qc  =  [0.082  X  0.24  +  0.0002  X  0.39][260°  -  65°]  =  3.86 

Qr  =  [0.082  X  0.24][25  -  (  -  99)]  =  2.44 

2.44 
Coefficient  of  performance  without  friction  =  g^g-  =  2.10. 

778" 
2.44 
Coefficient  of  performance  with  friction  = 


78 
2.44 
B.t.u.  of  refrigeration  per  ft.-lb.  of  work  =  TKO    —  0.00159. 


B.t.u.  per  h.p.-hr.  =  0.00159  X  33,000  X  60  =  3120. 

3120  X  24 
Tons  of  ice  melting  capacity  per  24  hrs.  per  h.p.  =  2000  X  144 

Assuming  3  Ibs.  of  coal  per  horse-power  hour: 


REFRIGERATION  423 

Tons  of  ice  melting  capacity  per  24  hrs.  per  Ib.  coal  =  0    '   0 ,   =  0.00361. 

o  X  ^Tt 
2.44 
Tons  of  ice  melting  capacity  per  cu.  ft.  =  1 44  x  OQQO  =  0.0000085. 

Gallons  of  cooling  water  per  cu.  ft.  of  piston  displacement  for  a  15°  rise 

O    Q£I 

in  water  temperature  =  1 ..    '  Q  QK  =  0.0305. 

1O   X  o.oO 

Gallons  per  ton  of  ice  melting  capacity  =  n" 0^00085  =  ^^. 

Relative   volume   of  expansion   cylinder  and    compression   cylinder  = 

T        364 

— -  =  -—    =  0.745.     (Air  removed  by  freezing  in  short  time) 

J.  i       "±00 

If  this  is  worked  out  without  moisture  the  following  results 
are  found: 

^4J^  =  0'082 
Tl  =  485°  abs.  =  25°  F. 

0.4 

T2  =  485(4) IA  =  720°  abs.  =  260°  F. 
Tz  =  65  +  460"  =  525°  abs. 

A.Q  P\ 

7%  =  525  X  ^  =  354°  abs.  =    -  106°  F. 

Wc  =  3600 

We  =  3600  X  HI  =  2630 

Net  work  =  1.10  X  3600  -  0.9  X  2630  =  1593  ft.-lbs. 
Qc  =  0.082  X  0.24[260°  -  65°]  =  3.84 
Qr  =  0.082  X  0.24[25°  -  (  -  106°)]  =  2.58 

2  58 
Coefficient  of  performance  =  ^7^  =  1.26. 


2  58 
B.t.u.  per  ft.-lb.  of  work  =  ^93=  0.00162. 

B.t.u.  of  refrigeration  per  h.p.-hr.  =  0.00162  X  33,000  X  60  =  3180. 

3180  X  24 
Tons  of  ice  melting  capacity  per  24  hrs.  per  h.p.  =  OQQQ  X  144  =  0-264. 

Assume  3  Ibs.  of  coal  per  h.p.-hr.: 

0  264 

Tons  of  ice  melting  capacity  per  24  hrs.  per  Ib.  coal  =  Q  \,  0/)  =  0.00368. 

^  o  X  ^ 

Tons  of  ice  melting  capacity  per  cu.  ft.  of  compression  cylinder  without 

2  58 
clearance  =  144  x  2QQO  =  0.000009. 

Gallons  of  cooling  water  per  cu.  ft.  of  piston  displacement  for  a  15°  rise 

3  84 
in  water  temperature  =  1  5  \x  g  35  =  0.0306. 


Gallons  per  ton  of  ice  melting  capacity  =  nTJAnnQQ  =  3400. 

Relative   volume   of  expansion   cylinder  and   compression    cylinder  = 

t       354 
i  ~  485 


424 


HEAT  ENGINEERING 


The  actual  volumes  displaced  by  the  pistons  will  be  increased 
by  clearance.  The  volume  computed  divided  by  the  clearance 
factor  will  give  the  actual  volume. 

The  use  of  air,  although  cheap,  has  the  great  objection  that 
the  volume  of  the  compressor  and  expander  are  large  for  a 
given  amount  of  refrigeration.  To  reduce  the  size  of  the 
machines,  ammonia  or  other  volatile  liquids  are  used  as  the 
media.  The  substances  used  are  NH3,  SO2  and  CO2  and  certain 
mixtures  such  as  Pictet  fluid  composed  of  97  per  cent,  of  S02 
and  3  per  cent,  of  CO2.  The  advantages  and  -disadvantages  of 
the  various  liquids  will  be  discussed  later. 

VAPOR  REFRIGERATING  MACHINES 

In  machines  using  these  volatile  liquids  the  heat  is  abstracted 
not  by  an  increase  in  temperature  of  the  refrigerating  medium 
but  by  the  evaporation  of  a  liquid.  The  underlying  principle 

of  these  machines  is  to  bring 
the  vapor  to  such  a  pressure 
that  the  temperature  of  lique- 
faction is  slightly  above  the  tem- 
perature of  a  water  supply,  and 
by  means  of  this  water  the  ab- 


FIG.     194. — Arrangement    of    the 
vapor  refrigerating  machinery. 


FIG.  195.— P-V  diagram  of  the 
vapor  refrigerating  machine.  Com- 
pressor and  expansive  cylinder. 


straction  of  heat  is  possible.  This  abstraction  condenses  the 
vapor  and  after  the  pressure  on  the  liquid  is  so  reduced  that 
the  temperature  of  vaporization  is  very  low,  the  liquid  may 
evaporate  by  the  heat  abstracted  from  a  space  at  a  low 
temperature.  The  temperature  of  cooling  water  fixes  the  upper 
pressure  and  the  temperature  of  the  refrigerated  space  fixes 
the  necessary  low  pressure.  It  is  merely  a  matter  of  pressure 
regulation  to  fix  the  temperature  limits. 


REFRIGERATION 


425 


It  is  seen  more 


The  apparatus  used  to  accomplish  this  result  is  shown  in  Fig. 
194.     Fig.  195  represents  the  PV  diagram  for  the  cycle. 

In  the  compressor  A  the  ammonia  or  other  vapor  is  com- 
pressed adiabatically  from  1  to  2.  This  will  superheat  the  am- 
monia vapor  if  dry  vapor  is  taken  into  the  cylinder,  while  if  there 
is  considerable  liquid  mixed  with  the  vapor  the  compression  will 
reduce  the  amount  of  this  liquid.  The  first  is  called  dry  com- 
pression and  the  second  wet  compression.  These  two  lines  are 
shown  in  Fig.  196  which  is  the  T-S  diagram  of  the  cycle. 
The  line  1-2  in  either  case  is  an  adiabatic. 
clearly  from  Fig.  196  that 
the  temperature  is  increased 
as  the  compression  takes 
place  so  that  when  the  com- 
pressed vapor  is  delivered 
from  the  compressor  into  the 
condensing  coil  or  condenser, 
B,  this  water  may  abstract 
heat  and  condense  the  vapor. 
This  occurs  from  2  to  3  in  Fig. 
196  while  in  Fig.  195  the  line 
2-3'  represents  the  discharge 
from  the  compressor  and  3'-3 
represents  the  volume  of  the  liquid.  At  times  the  liquid  is  cooled 
to  a  lower  temperature  by  passing  it  through  a  coil  cooled  by  the 
coldest  water.  This  is  shown  in  Fig.  196  by  3-3". 

If  now  the  liquid  be  allowed  to  expand  from  3  to  4  or  3"  to  4" 
on  an  adiabatic  in  an  expansion  cylinder  the  work  5345  would  be 
obtained.  This  amount  would  be  very  slight  and  the  return 
would  not  pay  for  the  complication.  Hence  the  pressure  is 
reduced  from  p2  to  pi  by  means  of  the  expansion  valve  C.  This 
throttling  action  results  in  a  value  of  the  heat  content  below  the 
valve  equal  to  that  above  the  valve. 

is  =  *Y  (54) 


FIG.  196      T-S  diagram  of   the  cycle 
of  the  vapor-refrigerating  machine. 


or 


This,  of  course,  reduces  the  refrigerating  effect  as 


and 


> 
> 


but  the  elimination  of  the  expansion  cylinder  has  made  the  appa- 
ratus simpler.     The  exact  loss  due  to  this  may  be  found  later. 


426  HEAT  ENGINEERING 

After  the  pressure  is  reduced  the  temperature  of  vaporization 
is  so  low  that  heat  will  be  abstracted  from  the  surrounding  sub- 
stances, even  though  they  be  at  a  low  temperature,  and  the  liquid 
vaporizes.  This  is  accomplished  in  the  refrigerating  coil  D. 
Here  the  evaporation  of  the  liquid  returns  either  dry  vapor  or 
wet  vapor  to  the  compressor  A  and  the  cycle  is  repeated. 

The  expressions  for  the  heat  on  the  various  lines  are  now 
computed. 

Heat  on  1-2  =  0 

Qc  =  heat  on  2-3  at  constant  pressure  =  i2  —  q's  (55) 

Qc  =  heat  on  2-3"  at  constant  pressure  =  i%  —  q'z"  (56) 

Heat  on  3-4,  3-4',  3"-4"  or  3"-4'"  =  0 

Qr  =  heat  on  4-1  =  ii  —  i*  or  ii  —  is  (57) 

But  i3  =  i\' 

.".  Qr  =  ii  -  is  or  ii  -  i3"  (58) 

Now  ii  is  the  expression  for  the  heat  content  at  either  Id  or 
Iw  and  iz  is  the  heat  content  at  2w  or  2d.  It  and  the  other  quan- 
tities may  be  found  in  ammonia  tables. 

i  =  q'  +  xr 

J7W. 
cpdt 
Tsat. 

The  expression  for  the  heat  added  on  a  constant  pressure  line 
from  a  to  6  is 

ib  -  ia  (59) 

as  is  given  under  Qr  and  Qc. 

The  equation  of  the  line  1—2  is  the  adiabatic 

s  =  constant 


s  =  s'i  +  -r  if  wet 
-/  1 

or  pvk  —  constant  if  dry. 

fc  =  1.33  for  superheated   ammonia.     Of   course  if  tables  for 
superheated  vapor  are  known,  the  formula 

Si   =   S2 

may  be  used  for  the  superheated  region  also.     This  is  really  the 
better  equation  even  in  the  superheated  region. 
The  work  required  is  the  algebraic  sum  of  the  heats 


REFRIGERATION 


427 


AW    =    Qr   ~    Qc    =     ~    [Qc   ~    Qr] 

—  AW  =  [i*  —  is']  -  [ii  -  i3'] 

=  iz  -  ii 
With  friction  -  AW  =  (1  +  f)(i*  -  ii) 


(60) 
(61) 


The  amounts  above  are  all  for  1  Ib.  of  ammonia  and  to  find  the 
amount  of  liquid,  per  minute,  hour  or  day,  the  amount  of  re- 
frigeration in  that  time  is  divided  by  the  refrigeration  per  pound. 
Thus  for  T  tons  of  refrigeration  per  day  the  weight  of  volatile 
liquid  per  minute  is 


The  horse-power  required  is 


(62) 


.-„ 


The  weight  of  cooling  water  per  minute  is 


where 


q'0  =  heat  of  liquid  for  condensing  water  at  outlet 
q\  =  heat  of  liquid  for  condensing  water  at  inlet. 


The  displacement  per  minute  of  the  compressor  is  given  by 

Mv"  Mv 


clearance  factor        clearance  factor 


The  displacement  of  the  cylinder  is 
found  by  computing  the  clearance 
factor  as  in  the  case  of  air  com- 
pressors. The  lines  of  expansion 
and  compression  are  of  the  same 
form.  This  is  shown  by  Fig.  197, 

the  actual  form  of  the  card.  If  V 
-  .  ,  .  ,  ,  ,  ,  .  . 

for  the  ammonia  on  the  lower  line  is 

known 


FIG.    197. — Actual  card  from 
a  compressor. 


The  refrigerative  effect  is  given  by 


In  many  cases  where  rooms  have  to  be  cooled,  the  ammonia 


428  HEAT  ENGINEERING 

or  air  is  not  sent  through  coils  in  the  room  but  heat  is  removed 
from  the  rooms  by  circulating  cold  brine  through  pipes.  The 
brine  is  cooled  by  the  air  or  ammonia  which  passes  through  a 
coil  in  a  tank  through  which  the  brine  is  pumped.  The  brine 
is  usually  a  solution  of  calcium  chloride  or  sodium  chloride. 
The  advantage  of  the  brine  system  over  the  direct  expansion 
system  is  the  fact  that  the  break  of  a  pipe  would  not  discharge 
the  ammonia  into  the  room  and  spoil  the  contents  and  also  the 
fact  that  the  compressor  may  be  shut  down  for  some  time  and 
the  cold  brine  stored  in  the  brine  tank  may  be  used  to  keep  the 
room  cool. 

These  formulae  may  be  used  for  any  volatile  liquid.  Even 
water  has  been  used  although  exceeding  low  pressures  must  be 
used  to  obtain  temperatures  below  32°  F.  A  problem  will  be 
computed  using  NHs,  SC>2  and  CO2  for  the  media  assuming  water 
at  50°  to  65°  F.  and  a  refrigerating  room  held  at  25°  F. 

In  this  problem  it  will  be  assumed  that  15°  is  the  necessary 
difference  of  temperature  for  heat  transfer  and  also  that  wet 
and  dry  compression  will  be  tried  for  the  ammonia.  For  dry 
compression  Xi  =  1  while  for  wet  compression  #2  =  1.  It  will 
be  assumed  that  " after  cooling"  reduces  T3"  to  50°  +  15°  =  65° 
F.  In  all  cases  the  volume  drawn  into  the  compressor  with 
1  per  cent,  clearance  will  be  assumed  to  be  1  cu.  ft. 
In  these  problems 

T4  =  Ti  -  25°  -  15°  +  460°  =  470°  abs. 

T2d  =  65°  +  15°  +  460°  =  540°  abs. 

Ty  =  50°  +  15°  +  460°  =  525° 

AMMONIA: 

Case  I. — Wet  compression. 
pi  =  38.02  Ibs. 
p2  =  153.90  Ibs. 
x2  =  1 


Xl= 

0.1025  +  0.9328  +  0.0483 


1.2017 
it  =  557 

ii  =  -  23.2  +  0.902  X  564.4  =  485.0 
i3'  =  36.5 

M  =  6^4  =  °'151 

Qe  =  0.151  [564.4  -  36.5]  =  78.6  B.t.u. 
Qr  =  0.151  [485  -  36.5]  =  67.8  B.t.u. 
AW  =  78.6  -  67.8  =  10.8  B.t.u. 
or  =  0.151  [564.4  -  485]  =  10.8  B.t.u. 


REFRIGERATION  429 


AW  with  friction  =  10.8  X  1.20  =  12.96 

67  8 
Refrigerating  effect  without  friction  =  ^^  =  6.23 


67.8 
Refrigerating  effect  with  friction    =  ^QQ  =  5.22 

A7  ft 

B.t.u.  per  ft.-lb.  of  work  =  12  96  x  778  =  °-00673 

B.t.u.  of  refrigeration    per  h.p.-hr.  =  0.00673  X  33,000  X  60  =  13,200 

13200  X  24 
Tons  of  refrigeration  per  h.p.  =  2000  X  144  = 

(Assume  3  Ibs.  of  coal  per  h.p.-hr.) 
Tons  of  refrigeration  per  Ib.  coal  =  Q    '   0/l   =  0.0153 

O     X     ^TC 

Tons  of  refrigeration  per  cu.  ft.    of  compressor   volume   taken   in  = 
67 '8         =  0.000235 


144  X  2000 
Gallons  of  cooling  water  per  cu.  ft.  of  compressor  volume  taken  in  for  15 

7ft  A 

rise  in  cooling  water  =  .,  -      V~^?  =  0.628 
lo  X  o.oo 

Gallons  per  ton  of  refrigeration  =  n  000235  =  2670 
Case  II.  —  Dry  compression. 
pi  =  38.02  Ibs. 
pz  =  153.90  Ibs. 
s2  =  Si  =  1.1534. 
This  is  s2  for  110°  F.  superheat. 
(This  may  be  checked  by 

0.25  /153  q\  0.25 

=664'ab, 


Deg.  superheat  =  664  -  540  =  124°  F.) 
i2  =  628.4  B.t.u. 

M  =  ^  =  0.136 

ii  =  541.2 
is'  =  36.5 

Qc  =  0.136  [628.4  -  36.5]  =  80.5 
Qr  =  0.136  [541.2  -  36.5]  =  68.6 
AW  =  80.5  -  68.6  =  11.9 
ATT  with  friction  =  11.9  X  1.20  =  14.3 

AC  A 

Refrigerating  effect  without  friction  ^^  =  5.76 

Aft  A 

Refrigerating  effect  with  friction  ^5  =  4.80 

14.  o 

AC  A 

B.t.u.  per  ft.-lb.  of  work  =  1A,    '  77Q  =  0.00617 

ll.o   X  /  i  o 

B.t.u.  of  refrigeration  per  h.p.-hr.  =  0.00617  X  33,000  X  60  =  12,200 

12200  X  24 
Tons  of  refrigeration  per  h.p.  =  2QOO  X  144  =  L°2 

Tons  of  refrigeration  per  Ib.  ccal  (assume  3  Ibs.  coal  per  h.p.-hr.) 
0.0142 


3  X  24 


430  HEAT  ENGINEERING 

Tons  of  refrigeration  per  cu.   ft.   of  compressor  volume  taken  in 

AC  A 

=  0.000238 


144  X  2000 

Gallons  of  cooling  water  per  cu.  ft.  of  compressor  volume  taken  in  for 

80  5 
15°  rise  in  cooling  water  =  ]g  x  o  35  =  0.644 

0.644 
Gallons  per  ton  of  refrigeration  =  Q  QQ0238  =  ^®® 

Carbon  Dioxide. — (Wet  compression). 
T4  =  Tl  =  470°  abs.  =  10°  F. 
T*d  =  540°  abs.  =  80°  F. 
T3  =  525°  abs.  =  65°  F. 
Pi  =  362.8  Ibs. 
p2  =  936  Ibs. 
*2  =  80 
s2  =  0.1511 

si  =  0.1511  =    -  0.0226  +  X!  0.2164 
zi  =  0.80 
ti  =    -  11.23  +  0.80  X  110.12 

=  76.86 
it  =  21.8 

»i  =  0.247  X  0.80  =  0.198 
M  =  5.05 

Qc  =  5.05[80  -  21.8]  =  294 
Qr  =  5.05[76.9  -  21.8]  =  278.4 
AW  =  5.05[80  -  76.9]  =  15.6 
AW  with  friction  =  15.6  X  1.50  =  23.4 

(Friction  has  been  increased  due  to  the  packing  made  necessary  by  the 
excessive  pressure.) 

278  4 
Refrigerating  effect  without  friction  =   j^r  =  17.8 

Refrigerating  effect  with  friction  =  -^^  =11.9 

070  4 

B.t.u.  per  ft.-lb.  work  =  234x778  =  °*0131 

B.t.u.  of  refrigeration  per  h.p.-hr.  =  0.0131  X  33,000  X  60  =  26,000 
Tons  of  refrigeration  per  h.p.  =  2000  v  144  =  2'17 

(Assume  3  Ibs.  coal  per  h.p.-hr.) 

2  17 
Tons  of  refrigeration  per  Ib.  coal  =  »    '   Oyl  =  0.0301 

O    X    ^Tt 

Tons  of  refrigeration  per  cu.   ft.   of  compressor  volume  taken  in  = 
0.000965 


144  X  2000 

Gallons  of  cooling  water  per  cu.  ft.  of  compressor  volume  taken  in  for  15C 

294 
rise  in  cooling  water  =  1{.  =  2.35 

1O   A  o.OO 

«  or 

Gallons  per  ton  of  refrigeration  =  Q  QQQ965  =  243° 


REFRIGERA  TION  43  1 


Sulphur  Dioxide.  —  (Wet  compression). 

T,  =  Tl  =  470°  abs.  =  10°  F. 

T.id  =  540°  abs.  =  80°  F. 

773  =  525°  abs.  =  65°  F. 

pi  =  13.5lbs. 

p2  =  59.65  Ibs. 

it  =  162.1 

S2  =  0.302 

si  =  0.302  =    -  0.1746  +  xi  0.3625 

xi  =  0.88 

ix  =    -  6.89  +  0.88  X  169.78  =  142.6 

is  =  10.96 

vi  =  0.88  X  5.96  =  5.25 

M  =  5^5  =  °'19 

Qc  =  0.19[162.1  -  10.96]  =  28.8 

Qr  =  0.19[142.6  -  10.96]  =  25.0 

AW  =  0.19[162.1  -  142.6]  =  3.71 

AW  with  friction  =  3.71  X  1.20  =  4.46 

25 
Refrigerating  effect  without  friction  =  x~yr  =  6.75 

25 
Refrigerating  effect  with  friction  =  j-^     =  5.61 


B.t.u.  per  ft.-lb.  work  =  A  AR  „  »nQ  =  0.00725 

4.4:0    X     •   «  O 

B.t.u.  per  h.p.-hr.  =  0.00725  X  33,000  X  60  =  14,350 
Tons  of  refrigeration  per  h.p.  =  OQQQ  y  144   =  1-195 
(Assume  3  Ibs.  coal  per  h.p.-hr.) 

Tons  of  refrigeration  per  Ib.  coal  =  »  '     2  ,  =  0.0167 

Tons  of  refrigeration  per  cu.   ft.  of  compressor  volume  taken   in  = 

OK 

=  0.000087 


144  X  2000 

Gallons  of  cooling  water  per  cu.  ft.  of  compressor  volume  taken  in  for  a 

28  8 
15°  rise  in  cooling  water  =  ^  ,  .'    og   =  0.23 

10  A  o.ou 

0  23 
Gallons  per  ton  of  refrigeration  =  Q  QQQQgf  =  ^640 


ABSORPTION  APPARATUS 

To  eliminate  as  far  as  possible  the  moving  parts  of  a  refriger- 
ating apparatus  and  to  utilize  waste  heat  from  other  machines  the 
absorption  type  of  refrigerating  machine  has  been  developed. 
In  this  the  high  pressure  is  produced  by  boiling  a  solution  of  NH3 
in  water  and  driving  off  the  NH3  by  a  steam  coil  at  such  a  rate  that 
the  pressure  is  sufficient  to  produce  a  temperature  of  condensa- 


432 


HEAT  ENGINEERING 


tion  above  the  temperature  of  the  supply  water.  The  low 
pressure  is  maintained  by  absorbing  the  vapor  in  water  at  a  rate 
to  produce  this  pressure. 

The  apparatus,  Fig.  199,  consists  of  a  generator  A  in  which  a 
steam  coil  B  supplies  enough  heat  to  warm  the  liquor  of  aqua 
ammonia  to  such  a  temperature  that  its  capacity  for  ammonia 
is  reduced  and  the  ammonia  is  driven  off.  This  gas  is  super- 
heated as  it  leaves  the  generator  and  to  use  some  of  the  heat 
above  saturation  (as  the  gas  must  finally  be  condensed)  it  is 
passed  over  a  series  of  baffle-plates  over  which  the  fresh  strong 
liquor  passes  on  its  way  to  the  generator  B.  The  part  of  the 
apparatus  C  is  called  the  analyzer.  The  function  of  this  is  to 


Cooling  Water  Outlet 
T  D  E 


Brine 
Outlet 


Condenser 


Strong 
Liquor 


FIG.  198. — Absorption  machine. 


reduce  the  superheat  in  the  discharge  gases.  As  the  gas  or  vapor 
of  ammonia  contains  some  water  vapor  which  would  interfere  with 
the  action  of  the  vapor  in  the  expansion  valve,  and  as  it  would  use 
some  of  the  refrigerating  capacity  when  it  absorbs  ammonia 
after  its  condensation,  a  rectifier  D  is  used  to  reduce  the  water 
content.  This  rectifier  cools  the  vapors  still  further.  The  heat 
is  absorbed  by  cold  water  or  by  the  strong  liquor  which  is  pumped 
from  the  absorber  through  tubes  in  the  rectifier  on  its  way  to 
the  generator.  This  substance  is  at  a  lower  temperature  than 
the  vapors,  and  although  low  enough  to  condense  most  of  the 
water  it  is  not  low  enough  to  condense  the  ammonia.  The 
condensed  water  is  caught  in  a  separator  E  and  as  it  absorbs 
ammonia,  the  aqua  liquor  thus  formed  is  sent  back  to  the  gen- 
erator. From  the  rectifier  the  dry  vapor  is  sent  into  the  con- 
denser F  where  the  ammonia  vapor  is  condensed  by  a  cold  water 
supply.  The  liquid  is  now  passed  through  the  expansion  valve 


REFRIGERATION  433 

G  and  enters  the  expansion  coil  of  the  brine  cooler  H  in  which 
the  liquid  is  evaporated  by  the  abstraction  of  heat  from  the 
surrounding  brine.  If  the  direct-expansion  system  is  used  this 
heat  is  drawn  from  the  room.  The  vapor  is  drawn  into  the 
absorber  /  by  the  weak  liquor  which  has  been  allowed  to  flow 
from  the  generator.  The  great  capacity  of  this  weak  liquor  for 
ammonia  produces  a  reduction  in  pressure  of  the  vapor.  This 
absorption  of  ammonia  generates  heat  and  a  coil  /  through  which 
cool  water  is  circulated  is  placed  in  the  absorber  to  keep  the 
temperature  at  a  low  point.  This  low  temperature  is  necessary 
to  keep  liquor  at  a  point  of  great  capacity  for  ammonia.  The 
absorbing  power  of  water  decreases  as  the  temperature  is  in- 
creased. The  strong  liquor  is  pumped  from  the  absorber  into 
the  generator  at  higher  pressure  by  the  pump  K.  To  save  some 
of  the  heat  it  is  passed  through  the  interchanger  L  in  which 
it  takes  heat  from  the  hot  weak  liquor  leaving  the  bottom  of 
the  generator.  The  weak  liquor  flows  to  the  absorber  /  in  which 
the  pressure  is  less  than  that  in  the  generator.  This  strong  liquor 
is  sometimes  first  passed  through  the  rectifier  pipes  as  it  is  cool 
enough  to  condense  the  water  vapor,  and  from  this  coil  it  passes 
to  the  interchanger  or  exchanger  L  and  then  the  strong  liquor 
is  discharged  into  the  analyzer  in  which  it  is  still  further  warmed 
by  the  superheated  vapors.  The  arrangement  in  the  figure,  how- 
ever, is  such  that  water  is  used  to  cool  the  analyzer.  The  strong 
liquor  finally  reaches  the  generator  and  the  boiling  drives  off 
the  ammonia. 

The  principle  of  the  absorption  machine  is  the  same  as  that  of 
the  commpression  machine.  The  vapor  is  put  under  such  a 
pressure  that  its  temperature  of  vaporization  is  above  that  of  a 
water  supply  and  this  water  may  remove  heat  and  condense 
the  ammonia;  after  this  it  is  placed  under  such  a  pressure  that 
heat  may  be  abstracted  from  a  region  of  low  temperature.  The 
pressure  is  produced  by  the  steam  coil  in  place  of  the  compressor 
and  the  low  pressure  is  maintained  by  absorption  in  place  of 
the  suction  of  the  compressor. 

PROPERTIES  OF  AQUA  AMMONIA 

The  action  of  ammonia  and  water  (known  as  aqua  ammonia) 
will  now  be  studied  to  form  a  basis  for  the  analysis  of  the  action 
of  this  machine. 

The  amount  of  ammonia  or  other  vapor  absorbed  by  a  pound 

28 


434  HEAT  ENGINEERING 

of  water  depends  on  the  pressure  and  temperature  of  the  water. 
Thus,  according  to  the  table  prepared  by  Lucke,  water  will  only 
absorb  3.1  per  cent,  of  its  weight  of  NH3  at  191°  F.  and  atmos- 
pheric pressure,  while  at  50  Ibs.  gauge  pressure  this  temperature 
for  the  same  concentration  could  be  raised  to  280°  F.  191.5°  F. 
is  the  temperature  limit  at  50  Ibs.  pressure  for  28  per  cent,  of 
absorption. 

The  relation  between  the  pressure,  per  cent,  of  ammonia  in 
a  solution  and  the  temperature  at  which  this  solution  will  give 
off  ammonia  or  rather  boil  has  been  determined  experimentally 
by  Mollier  and  although  plotted  in  curves  these  values  are 
represented  by  Maclntire  by  the  equation 

^  =  0.00466z  +  0.656  (67) 

1  sol. 

Taat.  =  temperature    of  saturation    of  ammonia  for  a  given 

pressure. 
Tgoi.  =  temperature  at  which  the  solution  boils. 

x  =  per  cent,   of  NH3  in   solution  =  per  cent,   of   con- 
centration. 
=  Ibs.  of  NH3  in  1  Ib.  of  solution. 

This  formula  holds  to  50  per  cent,  concentration.  Above 
that  concentration  the  formula  does  not  give  correct  results 
but  this  is  not  important  as  such  high  concentrations  are  rarely 
used. 

The  vapor  concentration  of  the  liquid  being  known  with  the 
temperature  of  boiling  (TsoL),  the  temperature  (Tsat.)  cor- 
responding to  the  boiling  pressure,  psat',  of  liquid  ammonia  is 
found.  This  gives  the  total  pressure  exerted  by  the  ammonia 
and  water  vapors  driven  off  from  the  liquor.  The  partial  pressures 
exerted  by  the  two  constituents  are  not  well  known  at  present. 
These  partial  pressures  are  proportional  to  the  number  of  mole- 
cules present  or  to  the  partial  volumes  of  each  constituent.  To 
know  how  these  molecules  come  off  from  the  solution  is  a  diffi- 
cult problem.  Lucke  quotes  values  of  partial  pressures  from 
Perman.  These  are  limited  to  140°  F.  and  to  concentrations 
from  2.5  to  22.5  per  cent,  by  weight  of  ammonia.  The  values 
indicate  the  manner  in  which  the  ammonia  vapor  pressure 
varies  with  the  water  vapor. 

Spangler  suggests  that  the  water  vapor  pressure  is  equal  to 
that  for  steam  at  the  temperature  given  multiplied  by  the  ratio 
of  number  of  the  molecules  of  water  vapor  to  the  total  number 


REFRIGERATION  435 

of  molecules  present.  He  assumes  that  the  concentration  of 
the  vapor  is  the  same  as  the  concentration  of  the  liquid.  Thus 
if  x  is  the  per  cent,  solution  which  is  ammonia  or  the  amount 
of  ammonia  in  100  Ibs.  of  liquor  the  ratio  of  the  number  of 
ammonia  molecules  and  water  molecules  is 

x  ,     100  -  x 

I7t(         18" 

Hence  if  p  is  the  steam  pressure  for  the  temperature  considered : 

100-  x 

18  1700  -  17  x 

P100  -  x        x^  '"  P     1700  +  x 
18  17 

=  partial  pressure  of  moisture.     (68) 

This  checks  with  the  results  of  Perman,  as  given  by  Lucke, 
with  a  close  degree  of  approximation. 

If  1  Ib.  of  ammonia  be  added  to  a  large  quantity  of  water 
it  is  found  that  893  B.t.u.  of  heat  will  be  liberated.  This 
is  called  the  heat  of  complete  absorption.  If  the  ammonia  is 
not  mixed  with  a  large  quantity  of  water  the  absorption  will 
be  partial.  If  strong  ammonia  liquor  is  added  to  water  it  is 
found  that  heat  is  developed  by  the  solution  of  this  in  water; 
the  strong  solution  acting  as  ammonia.  This  latter  heat  is  called 
the  heat  of  complete  dilution  and  the  difference  between  the 
heat  of  complete  absorption  and  that  of  complete  dilution  is 
known  as  the  heat  of  partial  absorption.  Complete  dilution 
occurs  when  the  ammonia  is  diluted  with  200  times  its  weight 
of  water. 

It  has  been  found  by  Berthelot  that  the  amount  of  heat  de- 
veloped when  a  liquor  of  aqua  ammonia  is  diluted  so  that  the 
water  content  is  200  times  the  weight  of  the  ammonia,  if  multi- 
plied by  the  weight  of  water  per  pound  of  ammonia  in  the 
original  solution  gives  a  constant  or  practically  constant  prod- 
uct, 142.5.  The  heat  of  complete  dilution  becomes 

142.5  142.5a 

^  :=  100  -  x  ~  100-s  (69) 

x 

x  =  per  cent,  weight  of  NH3  in  1  Ib.  of  mixture. 

Some  later  results  by  Thomsen  pertaining  to  much  greater 
dilutions  than  those  of  the  useable  experiments  of  Berthelot's 


436  HEAT  ENGINEERING 

indicate  that  142.5  may  be  used,  the  constant  varying  from  154 
for  19.27  parts  of  water,  to  148  for  29.36  parts  and  113  for  56.33 
parts. 

If  now  893  B.t.u.  is  the  heat  of  complete  absorption  and 

142  5x 

-  is  the  heat  of  complete  dilution 
1UU       x 

893  —  irkn  ''    -  =  heat  of  partial  absorption.  (70) 

1LMJ  —  X 

The  heat  of  partial  absorption  is  the  heat  liberated  when  1  Ib. 
of  ammonia  in  solution  is  added  to  enough  water  to  bring  it  to 
a  concentration  of  x. 

If  T7     Ibs.  of  ammonia  are  absorbed  by  —  -    —  Ibs.  of  water 


the  heat  developed  will  be 
<*  - 


-  x) 

If  now  ammonia  were  absorbed  to  change  the  concentration 
from  x  to  x',  the  amount  of  ammonia  then  present  in  the  water 
contained  in  the  original  pound  of  solution  of  strength  x  would  be 


_ 
100  A  100  -  x' 

This  may  be  seen  from  the  fact  that  the  amount  of  water  in 
1  Ib.  of  a  solution  of  strength  x  is  --JQQ  —  and  the  amount  of 
ammonia  per  pound  of  water  in  the  second  condition  is 


100  -  x' 
Hence  the  amount  of  ammonia  in  —  ~~  —  Ibs.  of  water  is 


100  -  x       __?•__          *L  v  10°  ~  x 
100      X  100  -  x'  ~  100       100  -  x' 

The  heat  generated  by  the  addition  of  this  amount  of  ammonia 
to  bring  the  solution  to  strength  x'  is 

*'    ,  ,  100  -  x  r  142.5s' 

ioo  x 


The  heat  developed  by  the  addition  of  ammonia  to  change 
the  strength  of  the  solution  from  x  to  x'  is  therefore 

*'         100  -  x  r  142.5*  I  _  _^r893  _     142^1 

y  ~  ioo  x  ioo  -  *'L893  "  100  -  *'J    ioor*      ioo  -xJ 

(74) 


REFRIGERATION  437 


The  weight  of  ammonia  added  is 
100-  a 


X 


-  an  _  jc_ 

-  x'\      100  ~ 


lOO       100 

If  the  heat  is  divided  by  the  weight  the  result  will  be  the  heat 
per  pound  or 

,  =  893  -  ^-,  +  (75) 


This  is  the  amount  of  heat  liberated  when  1  Ib.  of  ammonia 
vapor  is  absorbed  by  a  liquor  of  concentration  x  to  change  the 
concentration  to  x'. 

The  above  expression  is  true  for  the  absorption  of  ammonia 
but  when  1  Ib.  of  ammonia  is  driven  from  a  solution  to  change 
the  concentration  xf  to  concentration  x  not  only  is  the  heat 
q  necessary  but  in  addition  to  this,  it  is  necessary  to  evaporate 
the  water  vapor  present  and  to  superheat  the  vapors.  The 
amounts  of  these  quantities  will  be  found  in  a  problem  later. 

The  density  of  the  liquor  of  aqua  ammonia  varies  with  the 
amount  of  ammonia  absorbed  decreasing  with  the  concentration. 
The  specific  gravity  is  given  for  various  concentrations  x  by 

4  Q  r  T2  rs     -i 

sp.  gr.  =  1  -  x —  +  — —  (76) 

The  strong  solution  would  be  lighter  and  would  stay  at  the  top 
were  it  not  for  diffusion. 

The  specific  heat  of  aqua  ammonia  will  be  assumed  to  be  unity. 

When  a  strong  liquor  of  strength  x'  is  changed  to  strength  x 
by  giving  off  1  Ib.  of  ammonia  the  number  of  pounds  of  strong 
solution,  y,  required  is  given  by 

x'  x 

^Too  ~  (y  ""  l'  loo  =  l 
i     -*- 

100         100  -  x 
V  =  ^r-       r=^n  (77) 


100       100 

The  weight  of  weak  solution  is  (y  —  1). 

The  above  discussion  gives  the  necessary  properties  of  aqua 
ammonia  and  to  apply  them  as  well  as  to  compute  the  heat  trans- 
fers in  the  various  parts  of  the  apparatus  a  problem  will  be 
solved. 


438  HEAT  ENGINEERING 

PROBLEM 

Find  the  amount  of  refrigeration  per  pound  of  ammonia  driven 
off  per  minute  from  25  per  cent,  solution  in  the  generator  if  the 
room  temperature  is  to  be  held  at  25°  F.  and  the  cooling  water 
varies  from  50°  to  65°  F.  How  much  steam  at  20  Ibs.  gauge 
pressure  and  x  =  0.9  would  be  required  per  minute  per  ton  of 
capacity?  Find  the  heat  transfer  of  the  various  parts  of  the 
apparatus  together  with  the  amount  of  water  used  and  a  heat 
balance. 

Temperatures  and  Pressures. 

Temperature  of  ammonia  in  condenser  with  a  15°  difference 
from  the  hottest  water  is  65°  +  15°  =  80°  F. 

Temperature  in  expansion  coil  with  15°  difference  of  tempera- 
ture is  25°  -  15°  =  10°  F. 

Temperature  of  steam    in  generator   (at  34.7  Ibs.   abs.)   = 
259°  F. 
Pressure  in  condenser 

Ammonia  pressure  (for  80°  F.)  =  153.9  Ibs. 
Steam  pressure  (for  80°  F.)         =      0.51b. 
Total  pressure  =154. 4  Ibs. 

Pressure  in  expansion  coil  (10°  F.)  =  38.02  Ibs. 
Pressure  in  absorber  [38.02  -  0.5  (assumed)]  =  37.52  Ibs. 
Temperature  of  saturation  of  ammonia  =  9.5°  F. 
Pressure  in  rectifier  [154.4  +  0.5]  =  154.9  Ibs. 
Temperature  limit  of  rectifier  (154.9  Ibs.)  =  80.4°  F. 
Pressure  in  analyzer  [154.9  +  0.5]  =  155.4  Ibs. 
Pressure  in  generator  [155.4  +  0.5]  =  155.9  Ibs. 
Temperature  limits  in  various  parts : 

Generator. — For  155.9  Ibs.  pressure  the  temperature  at  which  a 
25  per  cent,  solution  will  boil  is 

^'V  46°  =  0-00466  X  25  +  0.656 

1  801- 

T80l.  =  ~5~  =  700°  abs.  =  240°  F. 


This  is  the  lowest  temperature  possible  to  drive  off  the  am- 
monia with  25  per  cent,  concentration.  As  the  evaporation  is 
carried  on  the  concentration  becomes  less.  To  boil  the  liquor 
to  a  smaller  concentration  requires  a  temperature  higher  than 
240°  F.  The  heat  required  to  liberate  the  ammonia  from  the 


REFRIGERATION  439 

strong  liquor  reduces  the  temperature  of  the  weak  liquor  and 
makes  the  temperature  of  the  vapors  leaving  the  generator  that 
required  by  the  formula  for  the  strong  liquor.  This  will  be  the 
assumption  used  in  this  work. 

The  limiting  concentration  at  this  point  when  the  liquid 
has  time  to  be  brought  to  within  5°  of  the  temperature  of  steam, 
or  259  -  5  =  254°  F.,  is 

™  =  0.00466*  +  0.656 


254  +  460 

102 
x  =  T-^T  =  21.9  per  cent. 

This  gives  a  very  small  change  in  concentration.  Suppose  that 
the  gauge  pressure  is  raised  to  30  Ibs.  This  gives  a  temperature 
of  274°  F. 

°-00466*  +  °'656 

x  =  17.50  per  cent. 
This  will  be  used. 

Thus  the  pressure  in  the  condenser  fixes  the  pressure  in  the 
generator  and  this  with  the  concentration  fixes  the  temperature 
or  pressure  for  the  steam  used  to  boil  solutions. 

The  temperature  limit  of  the  absorber  is  given  by 

95-+-  460 
'——-    -  =  0.00466  X  25  +  0.656 

J-sol. 

469.5 
0.773 

Suppose  this  is  kept  at  145°  F. 

For  a  15°  difference  in  temperature  in  the  coils  of  the  absorber 
the  water  should  be  kept  at  130°  F.  Of  course  the  water  from 
the  condenser  could  be  used  here  as  its  temperature  is  not  above 
65°  F. 

Limiting  conditions  leaving  rectifier  are  given  by  the  tempera- 
ture of  condensation  of  the  ammonia.  This  is  80.4°  F.  so  the 
temperature  at  this  point  may  be  taken  at  90°  F.  The  possible 
concentration  at  90°  F.  and  pressure  154.9  Ibs.  is 


TSOL  =  TTT—  =  607°  abs.  =  147°  F. 


h  0.656 
x  =  70  per  cent. 
Having  these  limiting  concentrations,  the  amount  of  moisture 


440  HEAT  ENGINEERING 

at  the  various  points  may  be  found  except  at  the  discharge  from 
the  analyzer  because  at  that  point  the  temperature  is  not  known 
as  it  is  due  to  the  mixture  of  the  strong  liquor  and  the  liquor 
from  the  rectifier.  To  find  this,  the  amount  of  strong  liquor  to 
give  1  Ib.  of  ammonia  when  changed  from  25  per  cent,  concentra- 
tion to  17.5  per  cent,  concentration  is  computed.  The  formula 

100  -a 
y  =     xf  -x 

is  true  for  absorption  and  gives  11.00  Ibs.  for  y  for  the  conditions 
above. 

100-17.5 
y  =    25-17.5  =  1L°° 

When  ammonia  is  driven  off  some  steam  is  driven  with  the 
ammonia  and  for  this  reason  the  concentration  of  the  remaining 
liquor  is  made  greater.  To  find  the  value  of  y  the  conditions 
leaving  the  generator  must  be  known. 

Total  pressure  in  generator  ................................    155  .  9  Ibs. 

Temperature  in  generator  .................................   240°  F. 

Concentration  ...........................................   25  per  cent. 

(1700  _  17  V  2^\ 
1700  +  25  —  /  =  18'51bs- 
Partial  pressure  of  ammonia  ...............................    137.4  Ibs. 

Saturation  temperature  ammonia  ..........................   73  .  5°  F. 

Superheat  ..................................  ;  ............    166  .  5°  F. 

Heat  content  ammonia  ...................................   658  .  1 

Specific  volume  ammonia  .................................   3  .  07 

Saturation  temperature  steam  .............................   224°  F. 

Superheat  ............................................  ...    16°  F. 

Heat  content  steam  ......................................    1162 

Specific  volume  steam  ....................................   22  .  0 

o  ryj 

Weight  of  steam  with  1  Ib.  NH3  =  |^  =  .................   0.  1395 

The  value  of  y  may  now  be  found 

0.250  -  0.175  [y-  (l+  0.1395)]  =  1 

1  -  0.1995 
y=     -  =10'65 


This  is  weight  of  liquor  at  25  per  cent,  concentration  entering 
the  generator. 

The  weight  of  weak  liquor  leaving  is 

(10.65  -  1.1395)  =  9.5105  Ibs. 
of  17.5  per  cent,  concentration.     This  passes  to  the  absorber. 


REFRIGERATION  441 

The  amount  of  ammonia  absorbed  by  bringing  this  solution  to 
a  concentration  of  25  per  cent,  is  given  by 

100 

[9.5105  (1.00  -  0.175)  -==-  -  9.5105]  = 
*  o 

10.461  -  9.5105  =  0.9505 

This  is  the  amount  of  ammonia  absorbed  but  there  is  still  a 
further  amount  absorbed  by  the  water  in  the  rectifier  and  an- 
alyzer. That  leaving  the  rectifier  includes  a  further  amount, 
although  small,  which  is  taken  over  by  the  condensed  moisture 
formed  from  the  moisture  leaving  the  rectifier  and  passing  into 
the  condenser  and  from  it  into  the  expansion  coils  and  absorber. 

The  amount  of  water  vapor  leaving  the  rectifier  must  now  be 
found. 

Conditions  at  discharge  of  rectifier: 

Total  pressure  ..................................    154  .  9  Ibs. 

Temperature  ...................................   90°  F. 

Concentration  of  liquor  leaving  ...................   70  per  cent. 

Partial  steam  pressure  f"o.696  X  ^^^  ^  ^  7°1  =0.20  Ibs. 

Partial  ammonia  pressure  ........................  154  .  7  Ibs. 

Temperature  saturation  of  ammonia  ...............  80.3°  F. 

Superheat  ammonia  .............................  9  .  7°  F. 

Heat  content  ammonia  ..........................  564  .  4  B.t.u. 

Specific  volume  ammonia  ........................  1  .  99  cu.  ft. 

Temperature  saturation  of  steam  .................  53°  F. 

Superheat  steam  ................................  37°  F. 

Heat  content  ...................................  1100  B.t.u. 

Specific  volume  .................................  1640 

From  this  the  amount  of  moisture  leaving  per  pound  of  am- 
monia is 

-  0.0012 


The  amount  of  ammonia  absorbed  by  this  amount  of  water, 
if  condensed,  to  produce  a  25  per  cent,  concentration  is 

05 
0.0012  X  75  =  0.0004 

Hence  for  every  pound  of  ammonia  absorbed  0.9996  Ib.  will 
be  absorbed  by  the  weak  liquor  and  0.0004  Ib.  will  be  absorbed 
by  the  water  sent  in  from  condenser. 

Since  the  actual  weak  liquor  absorbs  0.9505  Ib.  per  pound  of 
ammonia,  the  total  weight  of  ammonia  absorbed  to  make  the 
strong  liquor  is 


442  HEAT  ENGINEERING 

5^-0  9509  Ib 
0.9996  " 

and  the  weight  of  strong  liquor  is 

•  '  I  '  .  •• 

9.5105  +  0.9505  +  0.9509  X  0.0016  =  10.4625 

This  liquor  at  145°  F.  is  passed  through  the  interchanger  which 
is  supplied  with  9.5105  Ibs.  of  weak  liquor  at  240°  F.  The  heat 
given  up  by  this  liquor  in  reducing  its  temperature  to  150°  F. 
by  the  countercurrent  interchangers  is  equal  to 

Q  =  9.5105  [240  -  150]  X  1  =  856  B.t.u. 

This  assumes  that  the  specific  heat  of  the  liquor  is  1. 

If  a  20  per  cent,  radiation  loss  is  assumed  from  the  interchanger 
the  temperature  of  the  strong  liquor  leaving  this  apparatus  and 
entering  the  analyzer  is 


This  strong  liquor  is  mixed  with  a  small  amount  of  liquor-  of 
higher  concentration  but  at  90°  F.  from  the  rectifier.  This  will 
decrease  the  temperature  although  the  heat  developed  by  the 
solution  of  the  strong  liquor  will  increase  this  latter  temperature. 
The  net  effect  is  to  decrease  the  temperature  about  3^°.  The 
exact  amount  of  decrease  will  be  found  but  to  get  the  quantity 
of  liquor  from  the  rectifier  for  first  approximation  it  will  be  well 
to  assume  a  3J^°  drop  rather  than  no  drop  at  all. 

The  gases  from  the  analyzer  will  be  brought  to  the  temperature 
of  the  liquors  entering  the  top  of  the  analyzer  by  the  cooling  effect 
of  these  as  the  heat  of  superheat  is  much  less  than  the  heat  re- 
quired to  raise  the  temperature  of  the  liquor  to  that  at  the  desired 
discharge  into  the  generator  from  the  analyzer.  The  conditions 
of  the  vapors  leaving  the  analyzer  and  entering  the  rectifier  are 
fixed  by  the  temperature  and  pressure  at  this  point. 

Pressure  .........................................    155.4  Ibs. 

Temperature  assumed  .............................   207°  F. 

Concentration  f^y^:^  =  0.00466z  -f  0.6561  x    =33.2 

Partial  pressure  steam  [  13.3  X  17(f7~Q  ^  ^f'^]  =  8-72  Ibs. 
Partial  pressure  ammonia  ..........................    146.  7  Ibs. 


REFRIGERATION  443 

Temperature  saturation  ammonia 77 . 2°  F. 

Superheat  ammonia 129 . 8°  F. 

Heat  content  ammonia 638 . 9 

Specific  volume  ammonia 2.71 

Temperature  saturation  steam 187°  F. 

Superheat  steam 20°  F. 

Heat  content  steam 1149 . 3 

Specific  volume  steam 45 . 1 

2  71 
The  water  vapor  per  pound  of  ammonia  =  -r^r  =  0.06. 

4O.1 

Since  the  water  vapor  leaving  the  rectifier  per  pound  of  am- 
monia is  0.0012,  the  amount  of  water  removed  from  the  rectifier 
per  pound  of  ammonia  leaving  the  rectifier  is 

0.06  -  0.0012  =  0.0588 

In  addition  to  this  the  amount  of  moisture  associated  with 
the  ammonia  absorbed  is  also  condensed,  call  this  latter  M. 
The  concentration  of  the  liquor  leaving  the  rectifier  for  the  analy- 
zer is  70  per  cent.  Hence 

70 


[0.0588  +  M]  ~  |  0.06  =  M 
0.0588  X  -JQ  =  Jlf[l  -  gg 


M  =  0.00957 

Hence  the  moisture  condensed  per  pound  of  ammonia  entering 
the  condenser  is 

0.0588  +  0.0096  =  0.0684 

and  for  0.9505  Ib.  of  ammonia  this  is  0.065. 
The  ammonia  absorbed  by  this  is 

0.065  x||*  0.1517 

The  amount  of  liquor  passing  from  rectifier  is  then 
0.1517  +  0.065  =  0.2167  Ib. 

This  is  at  a  temperature  of  90°  F. 

On  mixing  with  10.4625  Ibs.  of  strong  liquor  at  210.5°  F.  the 
temperature  of  the  mixture  is  given  by: 

„,  _  10.4625  X  210.5  +  0.2167  X  90 
10.6792 

The  concentration  of  this  mixture  is  now  found. 


444  HEAT  ENGINEERING 

Ammonia  from  strong  liquor  =  10.4625  X  0.25  =   2  .  61572 

Ammonia  from  rectifier  liquor         =  =  0.15170 

Total  ammonia  .............................       2  .  76742 

2.76742 
Concentration  =  1Q  5790  =  25.913  per  cent. 

The  temperature  of  such  a  concentration  at  a  pressure   of 
155.4  Ibs.  is  given  by: 

=  0.00466  X  25.91  +  0.656 


1  sol- 

T80L  =  697°  abs.  =  237°  F. 

Since  the  temperature  of  the  mixture  is  below  this  point  the 
solution  would  remain  of  strength  given  were  it  not  for  the 
heat  of  solution  when  the  stronger  rectifier  liquor  is  diluted  in 
the  strong  liquor. 

To  change  10.4652  Ibs.  of  liquor  from  strength  25  per  cent. 
to  25.91  per  cent,  requires  an  amount  of  ammonia  equal  to 

10.4625  [0.2591  -  0.25]  =  0.0955 

This  should  also  equal  the  ammonia  lost  by  the  rectifier  liquor 
0.2167  [0.70  -  0.25913]  =  0.0955. 

The  mixture  of  a  strong  solution  in  a  weaker  solution  de- 
velops heat.  This  may  be  considered  as  the  difference  between 
the  heat  developed  when  the  weak  solution  is  made  strong  and 
that  required  to  weaken  the  strong  solution. 

Heat  = 


=  0.0955[g-|]l42.5 

-  0.0955  X  142;5[2.33  -  0.33] 
=  27.22 

This  heat  is  used  to  raise  the  temperature  of  the  mixture  and 
part  may  be  used  to  drive  off  ammonia  if  the  temperature  be- 
comes greater  than  the  temperature  of  solution  for  the  actual 
concentration  at  the  given  pressure.  The  temperature  rise  for 
27.22  B.t.u.  is 

27.22     _         0 
10.6792  =  2'55 

This  gives  208.05  +  2.55  =  210.55°  F. 


REFRIGERATION  445 

This  is  less  than  237°  F.  and  hence  there  is  no  evaporation. 
If  this  were  higher  it  would  be  well  to  assume  an  intermediate 
temperature  a  little  above  the  solution  temperature  for  the 
pressure  and  concentration  and  compute  the  concentration  for 
this  temperature.  Then  the  amount  of  ammonia  and  water 
vapor  driven  off  would  be  computed  together  with  the  heat 
required  to  do  this.  If  the  assumed  temperature  were  correct, 
this  heat  together  with  the  heat  to  raise  the  solution  from  208.7 
to  the  assumed  temperature  would  equal  27.22  B.t.u.  If  the 
sum  is  greater,  a  lower  temperature  is  assumed  and  the  quanti- 
ties computed.  After  several  assumed  temperatures  the  correct 
one  may  be  interpolated. 

In  any  case  the  final  answer  gives  the  temperature  at  the  dis- 
charge from  the  analyzer  and  the  second  value  of  condensation 
in  the  rectifier  may  be  found.  The  actual  conditions  at  discharge 
from  the  analyzer  into  the  rectifier  are  as  follows  : 

Pressure  .......................................  155  .  4  Ibs. 

Temperature  ...................................  210  .  25°  F. 

Concentration  ..................................  32  .  3 

Partial  steam  pressure  ...........................  9  .  39  Ibs. 

Partial  ammonia  pressure  ........................  146.01  Ibs. 

Temperature  saturation  ammonia  .................  76  .  9°  F. 

Superheat  ammonia  .............................  133  .  3°  F. 

Heat  content  ammonia  ..........................  640  .  9 

Specific  volume  ammonia  ........................  2  .  73 

Temperature  saturation  steam  .................  ...  190°  F. 

Superheat  steam  ................................  20  .  25°  F. 

Heat  content  steam  .............................  1151 

Specific  volume  steam  ...........................  42  .  4 

Water  vapor  per  Ib.  of  ammonia  ..................  0.0644 

Water  condensed  per  Ib.  of  ammonia  passing  into  condenser 
0.0644  -  0.0012  =  0.0632 

M  =  0.0644  [0.0632  +  M]  ^ 

<5U 

=  0.01117 

Total  moisture  =  0.0632  +  0.01117  =  0.0744  per  Ib. 
Moisture  for  0.9509  Ib.  =  0.0707 

Ammonia  absorbed  =  0.0707  X  T^  =  0.165 

Liquor  from  rectifier  =  0.165  +  0.0707  =  0.2357 

„  ,     .  x  10.4625  X  210.5  +  0.2357  X  90 

Temperature  of  mixture  =  -  1Q  -  =   207.84 


2  7807 
Concentration  of  mixture  =  1Q  6982  =  25.99  per  cent. 


446  HEAT  ENGINEERING 

Temperature  solution  =  696°  abs.  =  236°  F. 
Ammonia  taken  up  by  weak  solution 

10.4625  [0.2599  -  0.25]  =  0.1036 
Ammonia  given  up  by  strong  solution 

0.2357  [0.70  -  0.2599]  =  0.1037 

Heat  of  change  of  solution  0.1037  |^-  ^  1  142.5  =  29.55 

29  55 
Temperature  rise  =  IQ  gob 6  =  2-76°  F. 

Temperature  of  discharge  =  207.84  +  2.76  =  210.6°  F. 

This  is  below  236°  F.,  hence  there  is  no  further  change  of  con- 
centration and  the  condition  of  the  entrance  into  rectifier  of 
210.25  may  be  used  as  the  difference  in  temperature  does  not 
change  data. 

The  ammonia  entering  rectifier  is  therefore 

0.9509  +  0.165  =  1.1159 
The  vapor  entering  is  given  by  two  methods: 

Vapor  =  [0.0744  +  0.0012]  0.9509  =  0.0719 
Vapor  =  1.1159  X  0.0644  =  0.0740 
Mean  value  =  0.073 

The  liquor  at  temperature  210.25°  F.  drops  through  the 
analyzer  and  receives  heat  from  the  hot  vapors.  As  this  heat 
is  not  sufficient  to  heat  the  liquor  to  240°  F.,  it  will  be  assumed 
that  a  steam  coil  is  used  to  supply  the  necessary  heat  which 
will  be  computed.  The  liquor  is  assumed  to  leave  at  240°  F. 
The  conditions  for  this  are  given  for  exit  from  generator  on 
p.  440. 

Since  it  is  known  that  the  discharge  from  the  generator  is  1  Ib. 
of  ammonia  and  0.1395  Ib.  of  steam,  at  each  point  of  the  appa- 
ratus the  amount  of  substance  is  known. 

This  can  be  put  in  tabular  form: 

GENERATOR : 

Entering. — 10.65  Ibs.  liquor  of  25  per  cent,  concentration  at  240°  F. 
Leaving. — 1  Ib.  ammonia  vapor  at  240°  F. 

0.139  Ib.  water  vapor  at  240°  F. 

9.501  Ibs.  of  liquor  of  17.5  per  cent,  concentration  at  240°  F. 

ANALYZER  : 

Entering. — 1  Ib.  ammonia  vapor  at  240°  F. 

0.139  Ib.  of  water  vapor  at  240°  F. 

10.67  Ibs.  liquor  of  25.99  per  cent,  concentration  at  210.6°  F. 


REFRIGERATION  447 

Leaving. — 10.65  Ibs.  liquor  of  25  per  cent,  concentration  at  240°  F. 
1.116  Ibs.  of  ammonia  vapor  at  210.6°  F. 
0.074  Ib.  of  water  vapor  at  210.6°  F. 

RECTIFIER: 

Entering.— 1.116  Ibs.  of  ammonia  at  210.25°  F. 

0.074  Ib.  of  water  vapor  at  210.25°  F. 
Leaving. — 0.9509  Ib.  ammonia  vapor  at  90°  F. 

0.0011  Ib.  water  vapor  at  90°  F. 

0.236  Ib.  liquor  of  70  per  cent,  concentration  at  90°  F. 

CONDENSER  : 

Entering. — 0.9509  Ib.  ammonia  vapor  at  90°  F. 

0.0011  Ib.  of  water  vapor  at  90°  F. 
Leaving. — 0.9505  Ib.  of  ammonia  liquid  at  80°  F. 

0.0015  Ib.  liquor  of  strength  25  per  cent,  at  80°  F. 

EXPANSION  COIL: 

Entering.— 0.9505  Ib.  ammonia  liquid  at  80°  F. 

0.0015  Ib.  liquor  of  strength  25  per  cent,  at  80°  F. 
Leaving. — 0.9505  Ib.  ammonia  at  10°  F. 

0.0015  Ib.  water  vapor  (strength  25  per  cent,  assumed)  at  10°  F. 

ABSORBER: 

Entering. — 9.501  Ibs.  liquor  of  17.5  per  cent,  concentration  at  150°  F. 

0.9505  Ib.  ammonia  at  10°  F. 

0.0015  Ib.  water  vapor  at  10°  F. 
Leaving. — 10.463  Ibs.  liquor  of  25  per  cent,  concentration  at  145°  F. 

The  heat  interchange  for  each  part  is  found  by  the  difference 
between  the  intrinsic  energy  and  the  work  (or  heat  content)  at 
entrance  and  exit  plus  an  allowance  for  radiation. 

GENERATOR  : 

Entering.— Heat  of  liquid  of  10.65  Ibs. 

10.65[240  -  32]  =  2218 
Heat  of  solution  of  liquor 


1^893- 


-0.25  X  10.65|  893-  142.5  X  ^    =  -  2255 

/oj 

-     37 

Leaving. — Heat  of  1  Ib.  ammonia  =  1  X  658.1  =  658.1 

Heat  of  0.1395  Ib.  steam  =  0.1395  X  1162  =          162.0 
Heat  of  liquid  of  9.501  Ibs.  liquor  .  „ 

9.501  [240  -  32]  =  1975.0 

Heat  of  solution  of  liquor 


[893  - 


-0.175  X  9.501    893  -  142.5  X  =        -  1435.0 


1360.1 
Net  heat  to  be  supplied  =  1360.1  -  (-  37)  =  1397.1 


448  HEAT  ENGINEERING 

ANALYZER: 

Entering.  —  Heat  of  1  Ib.  ammonia  =  1  X  658.1  =  658.1 

Heat  of  0.1395  Ib.  steam  =  0.1395  X  1162  =  162.0 

Heat  of  liquid  of  liquor  =  10.67  [210.25  -  32]  =  1910.0 


[OCQQT 
893  -  142.5  X  7401]  =  -  2340.0 

390.1 
Leaving.—  Heat  of  liquid  of  10.65  Ibs.  =  10.65  [240  -  32]  =  2218 

[25~l 
893  -  142.5  X  75     =  -  2255 

Heat  of  1.116  Ibs.  ammonia  =  1.116  X  640.9  =  716.0 

Heat  of  0.074  Ib.  steam  =  0.074  X  1151  =  85.2 

764.2 
Net  amount  added  =  764.2  -  390.1  =  374.1 

Total  amount  added  in  generator  and  analyzer  allowing  20 
per  cent,  for  radiation  is 

120  [1397.1  +  374.1]  =  2125  B.t.u. 

Pounds  of  steam  required  at  30  Ibs.  gauge  and  x  =  0.9  is  given 
by 

Mat  -  0.9  X927.9  =  2'55  lbs' 
RECTIFIER: 

Entering.  —  Heat  of  1.116  lbs.  ammonia  =  716.0 

Heat  of  0.074  Ib.  steam  =  85.2 

801.2 

Leaving.  —  Heat  of  0.9509  Ib.  ammonia  =  0.9509  X  564.4  =  536.0 

Heat  of  0.0011  Ib.  steam  =  0.0011  X  1100  =  1.2 

Heat  of  liquid  of  0.236  Ib.  liquor  =  0.236  [90  -  32]  =  13.7 

Heat  of  solution  =  -  0.7  X  0.236   s93  -  142.5  X      1  =  -  92.5 


458.4 
Net  heat  abstracted  =  801.2  -  458.4  =  342.8  B.t.u. 

CONDENSER : 

Entering. — Heat  of  0.9509  Ib.  ammonia  =  536.0 

Heat  of  0.0011  Ib.  steam  =  1.2 

537.2 

Leaving.— Heat  of  liquid  of  0.9059  Ib.  ammonia  =  0.9509  X  53.6  =    51.0 
Heat  of  liquid  of  0.0011  Ib.  water  =  0.0011  X  48.1  =  0.0 

Heat  equivalent  of  ApV  =  0.0 

51.0 
Net  heat  abstracted  =  537.2  -51  =  486.2  B.t.u. 

EXPANSION  COIL: 

Entering. — Heat  of  ammonia  =  51.0 

Heat  of  water  =  0.0 

51.0 


REFRIGERATION  449 

Leaving. — Heat  of  ammonia  vapor  =  0.9505  X  541.2  =  514.0 

Heat  of  water  vapor  =  0.0015  [10  -  32]  =  -     0.0 

[25H 
893  -  142.5  X  75  I   =          -     3.4 

510.6 

(Work  of  liquid  neglected.) 

Heat  abstracted  510.6  -  51  =  459.6 

Tons  of  refrigeration  per  pound  of  vapor  per  minute: 

=  2.3  tons 


ABSORBER: 

Entering.— Heat  of  liquid  of  liquor  =  9.501  [150  -  32]  =  1120.0 

Heat  of  solution  of  liquor  =  -  0.175  X  9.501[862.7]  =  -  1435.0 
Heat  of  ammonia  =  514.0 

Heat  of  liquor  =  —3.4 

195.6 

Leaving.— Heat  of  liquid  of  liquor  =  10.463  [145  -  32]  =  1180 

Heat  of  solution  =  -  0.25  X  10.463  [845.5]    =  -2215 

-  1035 
Net  heat  abstracted  =  195.6  -  (-  1035)  =  1230.6 

INTERCHANGER : 

Entering.— Heat  of  liquid  of  strong  liquor  =  10.463  [145  -  32]  =     1180.0 
Heat  of  liquid  of  weak  liquor  =  9.501  [240  —  32]  =          1975.0 


Heat  equivalent  of  work  =  ^  I  155.4  -  37.5J  X 


144X10.463 
62.5  X  0.913 


3157.6 


43    /  252          353    \ 

[Density  =  1  -  ^  (25  -  ^  +  ^)  -  0.913] 


Leaving.—  Heat  of  liquid  of    strong  liquor  =  10.463  [210.5  -  32]  =  1865 
Heat  of  liquid  of  weak  liquor  =    9.501  [150  -  32]  =  1120 

2985 
Heat    radiated  =  3157.6  -  2985  =  172.6  B.t.u. 

2  6^ 
The  work  =  2.65  B.t.u  =     '  42  X  2  =  0.125  h.p.  for  1  Ib.  of  ammonia 

per  minute  with  50  per  cent,  efficiency. 


29 


450  HEAT  ENGINEERING 

HEAT  BALANCE: 


Taken  in  Given  out 

1397. I1 


Generator . . 
Analyzer. . . 


374.1 
354.0 


354.0 


Rectifier  .............  ..............  342.8 

Condenser  .........  .....  .....   486.2 

Expansion  coil  ....................  459  .  6 

Absorber  ......................  .  .  i|  i  .............  1230.6 

Pump  ..........................  :;..  2.65   . 

Exchanger  ..............  ......   172.6 

Total  ..........................  2587.4     ........  2586.2 

The  cooling  water  entering  the  condenser  at  50°  F.  is  raised  to 
65°  F.  This  could  then  go  to  the  rectifier  where  its  temperature 
could  be  raised  to  75°  F.  and  finally  to  the  absorber  where  the 
temperature  could  be  increased  to  130°  F.  The  weights  of  water 
would  then  be: 

Wt.  for  condenser  per  1  Ib.  ammonia  liberated   =     .  '       =  32.2 

342  8 
Wt.  for  rectifier  per  1  Ib.  ammonia  liberated  =         '      =  34.3 


6 
Wt.  for  absorber  per  1  Ib.  ammonia  liberated  =  —  ^.~-  =  22.4 

The  amount  of  water  actually  needed  is  34.3  Ibs. 

The  surface  required  in  these  different  sections  is  found  by 
methods  of  Chapter  III. 

The  other  data  computed  for  the  other  refrigerating  machines 
will  be  computed  for  comparison. 

Tons  of  refrigeration  per  Ib.  ammonia  =  OAAA  v  144  =  0-0^159 

459  6X8 
Tons  of  ice  melting  capacity  per  Ib.  coal  =  o  55  v  2000  X  144  =  0>00^ 

(Assuming  8  Ibs.  of  steam  per  Ib.) 
Tons  of  ice  melting  capacity  per  Ib.  steam  =  -^  —  =  0.0006 

120 
The   brine  pump  would  require    0.125  X  -QQ   =  0.25  Ib.    of 

steam  per  minute  for  459.6  B.t.u.  of  refrigeration  or  2.55  Ibs.  of 
steam  in  generator.  Hence  the  exhaust  from  the  pump  can 
be  used  easily  for  part  of  the  steam  supply. 

Gallons  of  cooling  water  with  15°  rise  per  ton  of  refrigeration  = 

34.3 
8.35  X  0.00159 


REFRIGERATION 


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452  HEAT  ENGINEERING 

From  the  above  it  is  seen  that  C02  gives  a  much  larger  capacity 
per  cubic  foot  and  is  much  better  for  efficiency  as  shown  by  tons 
of  refrigeration  per  pound  of  coal  or  steam.  The  air  machines 
have  small  capacities  per  cubic  foot  and  are  very  inefficient.  The 
absorption  machine  for  the  conditions  shown  is  not  as  efficient 
as  the  compression  apparatus  although  if  waste  steam  may  be 
used  this  has  a  great  value.  The  high  pressure  necessary  on 
the  CC>2  machine  is  the  objectionable  feature  of  this. 

TOPICS 

Topic  1. — Sketch  and  describe  the  action  of  an  air  refrigerating  machine. 
Give  expressions  for  the  heat  on  each  line.  Explain  what  is  meant  by  per- 
formance or  refrigerative  effect. 

Topic  2. — Sketch  and  describe  the  action  of  a  refrigerating  machine  using  a 
volatile  fluid  and  its  vapor.  Sketch  the  T-S  diagram  and  from  it  give  the 
expressions  for  the  heat  on  each  line.  Explain  what  is  meant  by  perform- 
ance. Explain  what  is  meant  by  a  ton  of  refrigeration. 

Topic  3. — -Sketch  and  explain  action  of  the  absorption  refrigerating  ma- 
chine. Give  the  function  of  each  part  of  this  apparatus. 

Topic  4. — Explain  method  of  finding  the  temperatures  and  pressures  of  the 
various  points  on  the  air  cycle  and  ammonia  cycle  of  a  refrigerating  machine. 
What  is  a  dense  air  machine?  Give  the  expressions  for  work,  heat  removed 
by  the  cooling  water  in  each  case  and  heat  removed  from  the  refrigerator. 

Topic  5. — What  is  the  effect  of  clearance?  Show  this  completely.  What 
is  the  effect  of  friction?  Give  the  expressions  for  the  power  necessary  for  air 
and  ammonia  machines  per  pound  of  substance  used.  How  is  the  volume  of 
the  compressor  found?  That  of  the  expander? 

Topic  6. — What  is  the  effect  of  moisture  in  an  air  machine?  Derive  an 
expression  for  the  work  in  the  expander  when  moisture  is  present  and  when 
no  moisture  is  present.  Derive  expressions  for  the  work  of  the  compressor 
with  and  without  moisture.  From  the  expressions  above  write  the  expres- 
sions for  work  with  and  without  friction. 

Topic  7. — -What  vapors  are  used  for  refrigerating  machines  ?  What  are  the 
advantages  of  one  over  the  other?  What  is  meant  by  wet  and  dry  compres- 
sion? Which  is  the  better?  Why?  What  is  the  clearance  factor?  Sketch 
T-S  diagrams  for  each  and  explain  action.  Why  is  the  expansion  valve  used  ? 
Write  the  expressions  for  work,  heat  and  refrigerating  effect. 

Topic  8. — What  is  aqua  ammonia?  What  is  the  effect  of  adding  ammonia 
to  water?  What  is  the  effect  of  adding  aqua  ammonia  to  water?  What  fixes 
the  pressure  at  which  ammonia  will  be  given  off  from  aqua  ammonia  at  a 
certain  temperature?  What  fixes  the  partial  pressures  of  the  ammonia  and 
water  vapors  discharging  from  this  solution?  What  is  meant  by  per  cent, 
ammonia  or  per  cent,  concentration? 

Topic  9. — What  is  the  heat  of  complete  dilution?  Complete  absorption? 
Partial  absorption?  What  is  manner  of  variation  of  the  density  of  aqua 
ammonia?  What  value  is  assumed  for  the  specific  heat  of  aqua  ammonia? 


REFRIGERATION  453 

Topic  10. — Outline  the  analysis  of  the  action  of  the  absorption  machine. 
What  data  is  assumed  and  what  does  this  fix? 


PROBLEMS 

Problem  1. — An  air  machine  is  closed  and  operates  between  40  Ibs.  gauge 
and  200  Ibs.  gauge  with  cooling  water  at  60°  F.  to  75°  F.  and  a  room  held  at 
5°  F.  Find  the  temperatures  at  the  various  corners.  Find  the  horse-power 
per  ton  of  refrigeration  with  10  per  cent,  friction  effect.  Find  the  cubic  feet 
of  displacement  per  minute  in  each  cylinder  and  the  amount  of  cooling  water 
per  minute  per  ton  of  refrigeration.  Clearance  5  per  cent,  on  each  cylinder. 

Problem  2. — An  air  machine  is  to  operate  under  conditions  of  Problem  1 
except  that  it  is  open  and  operates  to  45  Ibs.  gauge  pressure.  Find  the  quan- 
tities asked  for  in  Problem  1. 

Problem  3. — An  ammonia  machine  operates  with  cooling  water  from  60°  F. 
to  75°  F.  and  cools  a  room  with  direct  expansion  to  5°  F.  Find  the  pressures 
used.  Find  the  horse-power  per  ton  of  refrigeration  with  10  per  cent,  fric- 
tion effect  and  clearance  3  per  cent.  Find  the  amount  of  water  per  minute 
and  the  displacement  per  minute  for  1  ton.  (a)  Solve  this  with  wet  com- 
pression. (6)  Solve  this  with  dry  compression. 

Problem  4. — Solve  Problem  3  using  COa  as  the  medium. 

Problem  5. — Solve  Problem  3  using  SO2  as  the  medium. 

Problem  6. — A  25  per  cent,  solution  is  made  by  the  addition  of  ammonia 
to  a  10  per  cent,  solution  of  aqua  ammonia.  How  much  ammonia  has  been 
added  per  pound  of  original  solution?  What  is  the  density  of  each  solution? 
What  heat  is  developed  by  this  addition  of  ammonia? 

Problem  7. — A  25  per  cent,  solution  is  to  be  boiled  at  170  Ibs.  gauge 
pressure.  At  what  temperature  will  it  boil?  At  what  temperature  must 
this  be  heated  if  it  is  to  be  reduced  to  a  10  per  cent,  solution.  What  are  the 
partial  pressures  above  the  10  per  cent,  solution  in  this  case? 

Problem  8. — Find  the  amount  of  heat  per  pound  of  vapor  in  the  vapors 
coming  from  a  10  per  cent,  solution  of  aqua  ammonia  boiling  at  170  Ibs. 
gauge  pressure. 


INDEX 

Authors,  Authorities,  Inventors  and  Investigators 


Adams,  210 

American  Rad.  Co.,  101 
Andrew,  360 
Atkinson,  363 
Avogadro,  27 

Ball,  233 

Barnett,  359 

Barr,  209 

Barrus,  178 

Barsanti,  359 

Beau  de  Rochas,  360 

Bernouilli,  279 

Berthelot,  435 

Bertrand,  39 

Bisschop,  360 

Boltzmann,  73 

Boulvin,  189 

Box,  348 

Boyle,  26 

Brille,  80 

Brown,  359 

Buckingham,  314 

Buffalo  Forge  Co.,  101 

Callendar,  198,  204,  208 

Carcanagues,  80 

Carnot,  9,  11,  12,  13,  62,  170 

Carrier,  343 

Charles,  26 

Clark,  233 

Clausius,  167,  170 

Clayton,  210,  214 

Clerk,  360 

Coker,  369 

Corliss,  167 

Cotterill,  198 

Curtis,  302 

Dalby,  77,  80 
Dalton,  28 


Davis,  174,  317 
De  Laval,  301 
Dent,  156 
Diesel,  364 
Duchesne,  209 
Dulong,  73,  403 
Dupre  Hertz,  39 
Duhring,  356 

Fliegner,  60 
Foran,  335 

Giffard,  275 
Gilles,  360 

Goodenough,  16,  17,  25,  39,  44,  45, 
47,  215 

Hagemann,  99 
Hall,  205 
Halliday,  80 
Harn,  183 

Hausbrand,  99,  100,  356 
Hautefeuille,  359 
Heck,  199,  223 
Heinrich,  209 
Hilliwell,  360 
Huygens,  359 

Ingersoll  Rand  Co.,  120,  151 

Jelinek,  99 
Jordan,  80,  92 
Josse,  261 
Joule,  2,  37,  100 

Kelvin,  8,  36,  37 
Klebe,  44 
Kneass,  277,  279 
Knoblanch,  44,  45 
Kreisinger,  94,  108 


455 


456  INDEX 

Langen,  44,  360  Ray,  94,  108 

Le  Chatelier,  44  Reeve,  189 

Lenoir,  359  Reynolds,  77,  78,  167 

Linde,  44  Rice,  198 

Lucke,  73,  464  Richards,  156 

Ruggles,  348 
Mallard,  44 

Marks,  174,  198,  317  Sangster,  119 

Matteucci,  359  Savery,  167 

Maxwell,  24,  75  Scoble,  369 

Mclntire,  434  Ser,  80 

Me  Mullen,  96  Spangler,  177,  344,  396,  434 

Mollier,  45,  47,  80,  99,  100,  434              Stanton,  80,  91 

Moyer,  274,  292,  303  Stefan,  73 

Stodola,  269,  270,  271 

Napier,  60  Stone,  215 

Newcomen,  167  Street,  359 

Newton,  73,  359  Stumpf,  167,  226,  231 
Nicolson,  80,  89,  90,  91,  198,  204,  208      Sultzer,  167 
Nusselt,  75,  94 

Thomsen,  435 

Orrok,  97,  334,  335  Thurston,  198 

Otto,  360  Tyndall,  73 

Papin,  167,  359  Watt,  167 

Parsons,  305,  Webb,  232 

Perman,  434  Werner,  80 

Perry,  78,  198  Weymouth,  370 

Petit,  73  Willans,  167,  226 

Porter,  167  Worcester,  167 

Wright,  359 

Rankine,  60,  167,  170,  172,  215 

Rateau,  272,  304,  349  Zeuner,  215 


INDEX 

Subjects 


Absorber,  433 
Accumulators,  348 
Adiabatic,  2,  5,  16,  31,  54 
Adiabatic  action,  2 
Afterburning,  368 
Aftercooler,  120 
Air,  excess,  404 

motor,  151 

pump,  dry,  334 

transmission,  efficiency,  152 

loss,  152 

Allowance  for  turbines,  328 
Ammonia,  428 

aqua,  434 
Analysis,  Hirn's,  183 

proximate,  402 

T-S,  183 

ultimate,  402 
Analyzer,  432 

Aqua  ammonia,  concentration,  434 
density,  437 
heat  of  absorption,  435 
partial  pressure,  434 
properties,  433 
specific  heat,  437 
Ash  and  moisture,  free  basis,  402 
Availability,  1,  14,  62 

conditions  for,  14 

loss  of,  115 
Avogadro's  law,  27 
Axial  thrust,  300 

Back  pressure,  effect  of,  177 
Binary  engine,  261 
Blades,  number  of,  327 
Bleeding  engines,  255 
Blowing  engines,  119 

tubs,  119 

Bone-Schnable  boiler,  407 
Boyle's  law,  26 
British  thermal  unit,  1 


Calorimeters,  178 
Capacities,  thermal,  21 
Carbon  dioxide,  430 
Cards,  combined,  243 

computation  for,  243 

construction,  243 
Characteristic  equations,  19 
Charles'  law,  26 
Charts,  7-S,  47 

T-S,  47 

Coils  for  heating,  101 
Clearance,  effect,  124,  417 

effect  eliminated,  418 

factor,  125 

ratio,  125 

space,  125 

steam,  170,  184 
Closed  system,  412 
Combining  tube,  281 
Combustion,  375,  402,  403 

heat,  376 

surface,  406 

table,  377 
Complex  cycle,  38 
Compound  engine,  241 
Compounding    for    turbines,    pres- 
sures, 297 
velocity,  297 
Compression,  best  point,  230 

curve,  127 

dry,  425 

method,  128 

temperature,  127 

wet,  425 
Compressors,  118 

air,  118 
Condensation,  initial,  196 

maximum,  207 
Condenser,  ammonia,  98 

barometric,  337 

contraflo,  337 
457 


458 


INDEX 


Condenser  design,  338 

dry  tube,  337 

jet,  333 

steam,  97 

surface,  333 

uniflux,  337 
Conduction,  15,  72,  74 

coefficients,  95 

factors,  74 
Conductivity,  207 
Conservation  of  energy,  2 
Constant  steam  weight  curve,  217 

universal  gas,  27 
Consumption,  computation,  224 

curves,  222 

steam,  173,  210 
Convection,  72,  75 
Converging  nozzle,  273 
Cooling,  loss  from,  140 

towers,  341 
cost,  347 
size,  347 
Counter  flow,  82 

Criterion  for  exact  differentials,  4 
Critical  pressure,  267 

temperature,  49 
Cross  products,  37 
Curve,  construction,  218 

expansion,  215 

pressure  in  nozzle,  269 

steam  consumption,  222 
Curtis  turbine,  302,  311 

Dalton's  law,  28 

De  Laval  turbine,  301,  309 

Delivery  tube,  282 

shape,  282 

Determination  of  K,  86 
Diagrams,  combined,  240 

factor,  221 

I-S,  47 

Mollier,  47,  52 

temperature-entropy,  172,  175 
Differential  effect  of  heat,  43 

exact,  3 
Diffusivity,  207 
Dilution,  368 
Discharge,  59 
Displacement  of  air  compressor,  135 


Double  bottoms,  356 
Drip  pot,  object  of,  182 
Dry  compression,  425,  429 

basis,  402 

test,  187 

Dulong's  formula,  408 
Duty,  67 

Efficiency,  12 


air  standard,  362 

Carnot,  12,  14,  62 

change,  389 

conditions  for  maximum,  66 

cycle,  175 

equality,  12 

kinetic,  293 

maximums,  12,  13,  62,  65,  293 

mechanical,  63 

overall,  63 

practical,  63 

Rankine,  171,  173 

theoretical,  62 

transmission  of  heat,  103 

type,  63 

volumetric,  126 
Effects,    single,    double,    quadruple, 

350 

End  thrust,  306 
Energy,  conservation,  2 

high  grade,  1 

internal,  2 

intrinsic,  41 

Engine,  air,  displacement,  146 
power,  146 
work  on,  138 

and  turbine,  combined,  329 

compared,  309 
'binary,  261 

bleeding,  255 

heat,  62 

internal  combustion,  359 

Mietz  and  Weiss,  366 

multiple  expansion,  240 

Otto,  365 

receiver,  242 

regenerative,  256 

similar  to  turbine,  309 

size  of,  229 


INDEX 


459 


Engine  steam,  167 

steam  cycle,  170 

straight  flow,  170 

Stumpf,  226 

tests,  67,  68,  178 

uniflow,  226 

Woolf,  242 

work,  246 
Entropy,  16,  37,  43,  47,  51 

around  cycle,  18 

diagram,  19 

Equality  of  temperature  scales,  36 
Equation,  Bernouilli,  279 

characteristic,  19 

differential,  29 

fundamental,  21 
Equivalent  evaporation,  410 
Evaporation,  equivalent,  410 

factor,  410 

Evaporators,  98,  349,  350,  353 
Exact  differential,  3 
Expansion,  complete,  138 

curves,  construction  of,  218 

free,  172,  175 

incomplete,  138,  172,  175,  418 

line,  140,  215 
loss,  140 

ratio,  220 

Exploring  tube,  269 
Explosion  temperature,  389 
Exponent  value,  214 
External  work,  2,  31,  41 

Factor  diagram,  221 

of  evaporation,  410 

of  safety,  100 
Fan  blowers,  118 

power,  149 

Feedwater  heaters,  98 
First  law  of  thermodynamics,  2 
Flow,  counter  and  parallel,  82 

of  fluids,  56 
Free  air,  123 
Friction  factors,  292 
Fuels,  blast  furnace  gas,  374 

coal,  402 

crude  oil,  374 

gasoline,  374 

illuminating  gas,  370 


Fuels,  kerosene,  375 
lignite,  402 
natural  gas,  370 
peat,  402 
produce  gas,  371 
wood,  402 

Gas  composition,  376 

engine,    temperature,    entropy- 
diagram,  392 
test,  69,  70,  396 

producer,  371-373 
Generator,  432 
Governing,  air  compressors,  149 

gas  engines,  367 

steam  engines,  235 

multiple  expansion  engines,  254 
Graphical  representation  of  heat,  6 

Heat,  balance,  402,  450 

content,  37,  42,  47,  51 

engine,  62 

graphical  representation,  6 

internal,  41 

of  liquid,  40 

of  vaporization,  40 

on  line,  6,  7,  35 

on  path,  20,  42,  52 

transmission,  72,  75,  99,  103,  104 

efficiency,  103 
Hirn's  analysis,  183,  260 
High  grade  energy,   1 
Humphrey  compressor,  122 
Hyperbola,  rectangular,  55,  217 
Hydraulic  radius,  80 

compressor,  119,  122 

Ignition,  368 
Impact  coefficient,  284 
Indirect  heaters,  101 
Initial  condensation,  196 
Injector,  275 

capacity,  284 

density  of  discharge,  283 

details,  281 

double  jet,  284 

experiments,  277 

heat  equation,  280 

lifting,  284 


460 


INDEX 


Injector  number,  282 

overflowing  pressure,  285 
temperature,  285 

size,  282 

steam  required,  281 

theory,  280 
Interchanger,  433 
Intercooler,  120,  121,  130 

heat  removed,  133 

water  for,  133 

from,  134 
Internal  energy,  2 

heat  of  vaporization,  41 
Intrinsic  energy,  2,  42,  47,  51 

change,  35 

Irreversible  cycles,  15,  18 
Isodynamic,  6,  31,  54 

action,  2 
Isothermal,  7,  10,  31,  55 

Jacket,  121,  128,  253 

effect  of,  234 

heat  removed  by,  128 

saving  by,  129 

water  for,  130 
Jet,  force  of,  287 

impulse  of,  288 

reaction  of,  288 

work  of,  289 
Joule's  equivalent,  2 

Kinetic  energy  change,  265 

Lagging,  254 
Latent  heat,  21 
Law,  14 

Avogadro's,  27 

Boyle's,  26 

Charles',  26 

Dalton's,  28 

First  Law  of  Thermodynamics, 
2 

Second     Law     of     Thermody- 
namics, 14 

Stefan-Boltzmann,  73 
Leakage,  effect  of,  134 

factor,  125 

in  pipes,  143 

of  valves,  208 


Lines,  thermodynamics,  33,  50 

gas-engine  card,  392 
Locomobile,  235 

test  of,  68 

Logarithmic  diagrams,  153,  396 
Loss  curves,  274 

of  heat  186,  194 
Low-grade  energy,  1 

Mats,  341,  347 

Maxwell's  equations,  24 
Mean  effective  pressure,  219 
Missing  quantity,  196 
Mixtures,  gases,  28 
Molecular  weights,  29 
Moisture,  allowances,  328 

effect,  418 
Motors,  air,  151 
Mouth  of  nozzle,  268 
Mouthpiece,  272 
Multistaging,  130 

reasons  for,  135 

N-value  of,  214 
Nozzle,  265,  268,  271 

converging,  273 

spray,  348 

steam,  281 

velocity,  278 

Open  system,  412 
Orifices,  271,  272 
Overexpansion,  268 

Parallel  flow,  82 

Parson's  turbine,  305,  324 

Partitions,  heat  transmitted,  104 

Paths,  20 

Perfect  gases,  19,  26 

Performance,  method  of  reporting,  67 

Polytropic,  32 

Ponds,  cooling,  348 

Potentials,  thermodynamics,  24 

Power  of  motor,  126 

Preheater,  152 . 

Preheating,  gain  from,  145 

Pressure,  allowances,  328 

change,  effect  of,  176,  177 

critical,  267 


INDEX 


461 


Pressure  drop  in  pipes,  141 

intermediate,  130 

mean  effective,  219 

partial,  28,  334 

and  volume  of  steam,  225 
Producer,  gas,  371,  373 
Properties  of  aqua  ammonia,  433 
Proximate  analysis,  402 
Pumping  engine,  test,  69,  258 

Quadruple  expansion  engine,  241 
Quality,  40 

Radiation,  72 

factors,  74 
Radiators,  100 
Rateau,  accumulator,  349 

orifice,  272 

turbine,  304,  323 
Ratio  of  expansion,  220 
Receiver  engine,  242 
Rectangular  hyperbola,  55,  217 
Rectifier,  432 

Refrigerating  machines,  absorption, 
431 

air,  411 

dense  air,  416 

displacement,  414,  427 

efficiency,  413 

open  and  closed,  412 

power,  427 

temperature  range,  413 

vapor,  424 
Refrigeration,  tons  of,  414 

power  required,  414 
Refrigerative  effect,  427 

engines,  256 

performance,  413 
Refrigerator,  9 
Reheaters,  253 
Reheat  factor,  312,  323,  324 
Relations,  differential,  23 

trigonometric,  291 
Relative  humidity,  342 
Reversible  action,  10 
Rieman  steam  shock,  272 
Rotary  blowers,  119 

Safety  factor,  100 
Saturated  steam,  20 
Scale  of  temperature,  7 


Second  law  of  thermodynamics,  14 

Shock,  steam,  272 

Simple  cycle,  37 

Solutions,  dilute,  99 

Source,  9 

Specific  heat,  21,  30 

superheated  steam,  30 
Speed,  233 

Stage  compressors,  119 
Steam,  20,  39 

consumption,  173,  210 

curves,  222 

turbines,  287 

velocity,  278 

weight  curve,  217 

working,  185 

Stefan-Boltzmann  law,  73 
Still,  Hodges,  353 
Superheat  allowances,  328 

effect  of,  176,  177,  234 
Superheated  vapor,  44 
Sulphur  dioxide,  431 

Taylor  hydraulic  compressor,  122 
Temperature,  after  compression,  127 

after  explosion,  389 

boiling,  356 

critical,  49 

entropy  analysis,  188 
diagram,  392 

gas-engine  card,  364,  384 

gradient,  76 

intermediate,  133 

Kelvin's  absolute  scale,  8 

mean  difference,  81 

scale,  7 

equality,  36 
Tests,  analyses,  183,  188 

engines,  178 

gas  engines,  396 

pumping  engine,  258 
Thermal  unit,  British,  1 

capacities,  21 

relations  of,  22 
Thermodynamic  lines,  33 
Thermodymanics,  fundamental,  1 
Thermometers,  constant  immersion, 

181 
Throat  of  nozzle,  268 


462 


INDEX 


Throttling  action,  58,  123,  425 

loss  from,  144 
Thrust,  axial,  300 

end,  306 

Tons  of  refrigeration,  414 
Total  heat,  41 
Transmission  constants,  105 

heat,  72 

to  boiling  water,  99 
Trigonometric  relations,  291 
Triple  expansion,  241 
Tube  exploring,  269 
Tubs,  blowing,  119 
Turbine,  compared  with  engine,  309 

combined  with  engine,  329 

computations,  309 

Curtis,  302      • 

deLaval,  301 

efficiency,  293,  307 

low  pressure,  329 

maximum  efficiency,  293 

mixed  flow,  329 

Parsons,  305 

pressure,  291 

Rateau,  304 

steam,  167,  287 

test,  69 

velocity,  291 
Turbo  compressors,  118 

Ultimate  analysis,  402 
Underexpansion,  268 
Universal  gas  constant,  27 


Vacuum  allowance,  328 
Valves,  air,  136 

leakage,  208 
Vapor,  39 

Variables,  independent,  20 
Velocity,  actual,  289 

blade,  289 

discharge,  59,  265 

factors,  292 

injector,  278 

relative,  289 

whirl,  290,  292 
Vento  heaters,  101 
Viscous  liquids,  99 
Volume    and    pressure    of    steam, 

225 

Volumes,  partial,  28 
Volumetric  efficiency,  126 


Walls,  effect  of,  204,  369 
heat  transmission,  104 

Water  velocity,  278 

Weights,  molecular,  29 

Wet  compression,  425,  428 

Woolf  engine,  242 

Work,  31 

compression,  123 
multiple  expansion,  246 

Working  steam,  170,  185 


Y,  value  of,  266 


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